Elementa

Euclid

Elementa

Euclid · Elem 6 · perseus-eng2 · More by Euclid

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Original / primary: Perseus Eng2

6.def.1.p1548

Similar rectilineal figures are such as have their angles severally equal and the sides about the equal angles proportional.

6.def.2.p1549

[Reciprocally related figures. See note.]

6.def.3.p1550

A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less.

6.def.4.p1551

The height of any figure is the perpendicular drawn from the vertex to the base.

6.prop.1.p1552

Triangles and parallelograms which are under the same height are to one another as their bases.

6.prop.1.p1553

Let ABC, ACD be triangles and EC, CF parallelograms under the same height; I say that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF.

6.prop.1.p1554

For let BD be produced in both directions to the points H, L and let [any number of straight lines] BG, GH be made equal to the base BC, and any number of straight lines DK, KL equal to the base CD; let AG, AH, AK, AL be joined.

6.prop.1.p1555

Then, since CB, BG, GH are equal to one another, the triangles ABC, AGB, AHG are also equal to one another. [I. 38]

6.prop.1.p1556

Therefore, whatever multiple the base HC is of the base BC, that multiple also is the triangle AHC of the triangle ABC.

6.prop.1.p1557

For the same reason, whatever multiple the base LC is of the base CD, that multiple also is the triangle ALC of the triangle ACD; and, if the base HC is equal to the base CL, the triangle AHC is also equal to the triangle ACL, [I. 38] if the base HC is in excess of the base CL, the triangle AHC is also in excess of the triangle ACL, and, if less, less.

6.prop.1.p1558

Thus, there being four magnitudes, two bases BC, CD and two triangles ABC, ACD, equimultiples have been taken of the base BC and the triangle ABC, namely the base HC and the triangle AHC, and of the base CD and the triangle ADC other, chance, equimultiples, namely the base LC and the triangle ALC; and it has been proved that, if the base HC is in excess of the base CL, the triangle AHC is also in excess of the triangle ALC; if equal, equal; and, if less, less.

6.prop.1.p1559

Therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. [V. Def. 5]

6.prop.1.p1560

Next, since the parallelogram EC is double of the triangle ABC, [I. 41] and the parallelogram FC is double of the triangle ACD, while parts have the same ratio as the same multiples of them, [V. 15] therefore, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram FC.

6.prop.1.p1561

Since, then, it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD, and, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF, therefore also, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram FC. [V. 11]

6.prop.1.p1562

Therefore etc. Q. E. D.

6.prop.1.p1562

1

6.prop.1.p1562

2

6.prop.1.p1562

3

6.prop.2.p1563

If a straight line be drawn parallel to one of the sides of a triangle, it will cut the sides of the triangle proportionally; and, if the sides of the triangle be cut proportionally, the line joining the points of section will be parallel to the remaining side of the triangle.

6.prop.2.p1564

For let DE be drawn parallel to BC, one of the sides of the triangle ABC; I say that, as BD is to DA, so is CE to EA.

6.prop.2.p1565

For let BE, CD be joined.

6.prop.2.p1566

Therefore the triangle BDE is equal to the triangle CDE; for they are on the same base DE and in the same parallels DE, BC. [I. 38]

6.prop.2.p1567

And the triangle ADE is another area.

6.prop.2.p1568

But equals have the same ratio to the same; [V. 7] therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE.

6.prop.2.p1569

But, as the triangle BDE is to ADE, so is BD to DA; for, being under the same height, the perpendicular drawn from E to AB, they are to one another as their bases. [VI. 1]

6.prop.2.p1570

For the same reason also, as the triangle CDE is to ADE, so is CE to EA.

6.prop.2.p1571

Therefore also, as BD is to DA, so is CE to EA. [V. 11]

6.prop.2.p1572

Again, let the sides AB, AC of the triangle ABC be cut proportionally, so that, as BD is to DA, so is CE to EA; and let DE be joined.

6.prop.2.p1573

I say that DE is parallel to BC.

6.prop.2.p1574

For, with the same construction, since, as BD is to DA, so is CE to EA, but, as BD is to DA, so is the triangle BDE to the triangle ADE, and, as CE is to EA, so is the triangle CDE to the triangle ADE, [VI. 1] therefore also, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. [V. 11]

6.prop.2.p1575

Therefore each of the triangles BDE, CDE has the same ratio to ADE.

6.prop.2.p1576

Therefore the triangle BDE is equal to the triangle CDE; [V. 9] and they are on the same base DE.

6.prop.2.p1577

But equal triangles which are on the same base are also in the same parallels. [I. 39]

6.prop.2.p1578

Therefore DE is parallel to BC.

6.prop.2.p1579

Therefore etc. Q. E. D.

6.prop.3.p1580

If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle.

6.prop.3.p1581

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; I say that, as BD is to CD, so is BA to AC.

6.prop.3.p1582

For let CE be drawn through C parallel to DA, and let BA be carried through and meet it at E.

6.prop.3.p1583

Then, since the straight line AC falls upon the parallels AD, EC, the angle ACE is equal to the angle CAD. [I. 29]

6.prop.3.p1584

But the angle CAD is by hypothesis equal to the angle BAD; therefore the angle BAD is also equal to the angle ACE.

6.prop.3.p1585

Again, since the straight line BAE falls upon the parallels AD, EC, the exterior angle BAD is equal to the interior angle AEC. [I. 29]

6.prop.3.p1586

But the angle ACE was also proved equal to the angle BAD; therefore the angle ACE is also equal to the angle AEC, so that the side AE is also equal to the side AC. [I. 6]

6.prop.3.p1587

And, since AD has been drawn parallel to EC, one of the sides of the triangle BCE, therefore, proportionally, as BD is to DC, so is BA to AE.

6.prop.3.p1588

But AE is equal to AC; [VI. 2] therefore, as BD is to DC, so is BA to AC.

6.prop.3.p1589

Again, let BA be to AC as BD to DC, and let AD be joined; I say that the angle BAC has been bisected by the straight line A.D.

6.prop.3.p1590

For, with the same construction, since, as BD is to DC, so is BA to AC, and also, as BD is to DC, so is BA to AE : for AD has been drawn parallel to EC, one of the sides of the triangle BCE: [VI. 2] therefore also, as BA is to AC, so is BA to AE. [V. 11]

6.prop.3.p1591

Therefore AC is equal to AE, [V. 9] so that the angle AEC is also equal to the angle ACE. [I. 5]

6.prop.3.p1592

But the angle AEC is equal to the exterior angle BAD, [I. 29] and the angle ACE is equal to the alternate angle CAD; [id.] therefore the angle BAD is also equal to the angle CAD.

6.prop.3.p1593

Therefore the angle BAC has been bisected by the straight line AD.

6.prop.3.p1594

Therefore etc. Q. E. D.

6.prop.4.p1595

In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

6.prop.4.p1596

Let ABC, DCE be equiangular triangles having the angle ABC equal to the angle DCE, the angle BAC to the angle CDE, and further the angle ACB to the angle CED; I say that in the triangles ABC, DCE the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

6.prop.4.p1597

For let BC be placed in a straight line with CE.

6.prop.4.p1598

Then, since the angles ABC, ACB are less than two right angles, [I. 17] and the angle ACB is equal to the angle DEC, therefore the angles ABC, DEC are less than two right angles; therefore BA, ED, when produced, will meet. [I. Post. 5]

6.prop.4.p1599

Let them be produced and meet at F.

6.prop.4.p1600

Now, since the angle DCE is equal to the angle ABC, BF is parallel to CD. [I. 28]

6.prop.4.p1601

Again, since the angle ACB is equal to the angle DEC, AC is parallel to FE. [I. 28]

6.prop.4.p1602

Therefore FACD is a parallelogram; therefore FA is equal to DC, and AC to FD. [I. 34]

6.prop.4.p1603

And, since AC has been drawn parallel to FE, one side of the triangle FBE, therefore, as BA is to AF, so is BC to CE. [VI. 2]

6.prop.4.p1604

But AF is equal to CD; therefore, as BA is to CD, so is BC to CE, and alternately, as AB is to BC, so is DC to CE. [V. 16]

6.prop.4.p1605

Again, since CD is parallel to BF, therefore, as BC is to CE, so is FD to DE. [VI. 2]

6.prop.4.p1606

But FD is equal to AC; therefore, as BC is to CE, so is AC to DE, and alternately, as BC is to CA, so is CE to ED. [V. 16]

6.prop.4.p1607

Since then it was proved that, as AB is to BC, so is DC to CE, and, as BC is to CA, so is CE to ED; therefore, ex aequali, as BA is to AC, so is CD to DE. [V. 22]

6.prop.4.p1608

Therefore etc. Q. E. D.

6.prop.5.p1609

If two triangles have their sides proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.

6.prop.5.p1610

Let ABC, DEF be two triangles having their sides proportional, so that, as AB is to BC, so is DE to EF, as BC is to CA, so is EF to FD, and further, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and they will have those angles equal which the corresponding sides subtend, namely the angle ABC to the angle DEF, the angle BCA to the angle EFD, and further the angle BAC to the angle EDF.

6.prop.5.p1611

For on the straight line EF, and at the points E, F on it, let there be constructed the angle FEG equal to the angle ABC, and the angle EFG equal to the angle ACB; [I. 23] therefore the remaining angle at A is equal to the remaining angle at G. [I. 32]

6.prop.5.p1612

Therefore the triangle ABC is equiangular with the triangle GEF.

6.prop.5.p1613

Therefore in the triangles ABC, GEF the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles; [VI. 4] therefore, as AB is to BC, so is GE to EF.

6.prop.5.p1614

But, as AB is to BC, so by hypothesis is DE to EF; therefore, as DE is to EF, so is GE to EF. [V. 11]

6.prop.5.p1615

Therefore each of the straight lines DE, GE has the same ratio to EF; therefore DE is equal to GE. [V. 9]

6.prop.5.p1616

For the same reason DF is also equal to GF.

6.prop.5.p1617

Since then DE is equal to EG, and EF is common, the two sides DE, EF are equal to the two sides GE, EF; and the base DF is equal to the base FG; therefore the angle DEF is equal to the angle GEF, [I. 8] and the triangle DEF is equal to the triangle GEF, and the remaining angles are equal to the remaining angles, namely those which the equal sides subtend. [I. 4]

6.prop.5.p1618

Therefore the angle DFE is also equal to the angle GFE, and the angle EDF to the angle EGF.

6.prop.5.p1619

And, since the angle FED is equal to the angle GEF, while the angle GEF is equal to the angle ABC, therefore the angle ABC is also equal to the angle DEF.

6.prop.5.p1620

For the same reason the angle ACB is also equal to the angle DFE, and further, the angle at A to the angle at D; therefore the triangle ABC is equiangular with the triangle DEF.

6.prop.5.p1621

Therefore etc. Q. E. D.

6.prop.6.p1622

If two triangles have one angle equal to one angle and the sides about the equal angles proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.

6.prop.6.p1623

Let ABC, DEF be two triangles having one angle BAC equal to one angle EDF and the sides about the equal angles proportional, so that, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and will have the angle ABC equal to the angle DEF, and the angle ACB to the angle DFE.

6.prop.6.p1624

For on the straight line DF, and at the points D, F on it, let there be constructed the angle FDG equal to either of the angles BAC, EDF, and the angle DFG equal to the angle ACB; [I. 23] therefore the remaining angle at B is equal to the remaining angle at G. [I. 32]

6.prop.6.p1625

Therefore the triangle ABC is equiangular with the triangle DGF.

6.prop.6.p1626

Therefore, proportionally, as BA is to AC, so is GD to DF. [VI. 4]

6.prop.6.p1627

But, by hypothesis, as BA is to AC, so also is ED to DF; therefore also, as ED is to DF, so is GD to DF. [V. 11]

6.prop.6.p1628

Therefore ED is equal to DG; [V. 9] and DF is common; therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF; therefore the base EF is equal to the base GF, and the triangle DEF is equal to the triangle DGF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4]

6.prop.6.p1629

Therefore the angle DFG is equal to the angle DFE, and the angle DGF to the angle DEF.

6.prop.6.p1630

But the angle DFG is equal to the angle ACB; therefore the angle ACB is also equal to the angle DFE.

6.prop.6.p1631

And, by hypothesis, the angle BAC is also equal to the angle EDF; therefore the remaining angle at B is also equal to the remaining angle at E; [I. 32] therefore the triangle ABC is equiangular with the triangle DEF.

6.prop.6.p1632

Therefore etc. Q. E. D.

6.prop.7.p1633

If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, the triangles will be equiangular and will have those angles equal, the sides about which are proportional.

6.prop.7.p1634

Let ABC, DEF be two triangles having one angle equal to one angle, the angle BAC to the angle EDF, the sides about other angles ABC, DEF proportional, so that, as AB is to BC, so is DE to EF, and, first, each of the remaining angles at C, F less than a right angle; I say that the triangle ABC is equiangular with the triangle DEF, the angle ABC will be equal to the angle DEF, and the remaining angle, namely the angle at C, equal to the remaining angle, the angle at F.

6.prop.7.p1635

For, if the angle ABC is unequal to the angle DEF, one of them is greater.

6.prop.7.p1636

Let the angle ABC be greater; and on the straight line AB, and at the point B on it, let the angle ABG be constructed equal to the angle DEF. [I. 23]

6.prop.7.p1637

Then, since the angle A is equal to D, and the angle ABG to the angle DEF, therefore the remaining angle AGB is equal to the remaining angle DFE. [I. 32]

6.prop.7.p1638

Therefore the triangle ABG is equiangular with the triangle DEF.

6.prop.7.p1639

Therefore, as AB is to BG, so is DE to EF [VI. 4]

6.prop.7.p1640

But, as DE is to EF, so by hypothesis is AB to BC; therefore AB has the same ratio to each of the straight lines BC, BG; [V. 11] therefore BC is equal to BG, [V. 9] so that the angle at C is also equal to the angle BGC. [I. 5]

6.prop.7.p1641

But, by hypothesis, the angle at C is less than a right angle; therefore the angle BGC is also less than a right angle; so that the angle AGB adjacent to it is greater than a right angle. [I. 13]

6.prop.7.p1642

And it was proved equal to the angle at F; therefore the angle at F is also greater than a right angle.

6.prop.7.p1643

But it is by hypothesis less than a right angle : which is absurd.

6.prop.7.p1644

Therefore the angle ABC is not unequal to the angle DEF; therefore it is equal to it.

6.prop.7.p1645

But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32]

6.prop.7.p1646

Therefore the triangle ABC is equiangular with the triangle DEF.

6.prop.7.p1647

But, again, let each of the angles at C, F be supposed not less than a right angle; I say again that, in this case too, the triangle ABC is equiangular with the triangle DEF.

6.prop.7.p1648

For, with the same construction, we can prove similarly that BC is equal to BG; so that the angle at C is also equal to the angle BGC. [I. 5]

6.prop.7.p1649

But the angle at C is not less than a right angle; therefore neither is the angle BGC less than a right angle.

6.prop.7.p1650

Thus in the triangle BGC the two angles are not less than two right angles: which is impossible. [I. 17]

6.prop.7.p1651

Therefore, once more, the angle ABC is not unequal to the angle DEF; therefore it is equal to it.

6.prop.7.p1652

But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32]

6.prop.7.p1653

Therefore the triangle ABC is equiangular with the triangle DEF.

6.prop.7.p1654

Therefore etc. Q. E. D.

6.prop.8.p1655

If in a right-angled triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another.

6.prop.8.p1656

Let ABC be a right-angled triangle having the angle BAC right, and let AD be drawn from A perpendicular to BC; I say that each of the triangles ABD, ADC is similar to the whole ABC and, further, they are similar to one another.

6.prop.8.p1657

For, since the angle BAC is equal to the angle ADB, for each is right, and the angle at B is common to the two triangles ABC and ABD, therefore the remaining angle ACB is equal to the remaining angle BAD; [I. 32] therefore the triangle ABC is equiangular with the triangle ABD.

6.prop.8.p1658

Therefore, as BC which subtends the right angle in the triangle ABC is to BA which subtends the right angle in the triangle ABD, so is AB itself which subtends the angle at C in the triangle ABC to BD which subtends the equal angle BAD in the triangle ABD, and so also is AC to AD which subtends the angle at B common to the two triangles. [VI. 4]

6.prop.8.p1659

Therefore the triangle ABC is both equiangular to the triangle ABD and has the sides about the equal angles proportional.

6.prop.8.p1660

Therefore the triangle ABC is similar to the triangle ABD. [VI. Def. 1]

6.prop.8.p1661

Similarly we can prove that the triangle ABC is also similar to the triangle ADC; therefore each of the triangles ABD, ADC is similar to the whole ABC.

6.prop.8.p1662

I say next that the triangles ABD, ADC are also similar to one another.

6.prop.8.p1663

For, since the right angle BDA is equal to the right angle ADC, and moreover the angle BAD was also proved equal to the angle at C, therefore the remaining angle at B is also equal to the remaining angle DAC; [I. 32] therefore the triangle ABD is equiangular with the triangle ADC.

6.prop.8.p1664

Therefore, as BD which subtends the angle BAD in the triangle ABD is to DA which subtends the angle at C in the triangle ADC equal to the angle BAD, so is AD itself which subtends the angle at B in the triangle ABD to DC which subtends the angle DAC in the triangle ADC equal to the angle at B, and so also is BA to AC, these sides subtending the right angles; [VI. 4] therefore the triangle ABD is similar to the triangle ADC. [VI. Def. 1]

6.prop.8.p1665

Therefore etc.

6.prop.8.p1666

Porism. From this it is clear that, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the straight line so drawn is a mean proportional between the segments of the base. Q. E. D.

6.prop.9.p1667

From a given straight line to cut off a prescribed part.

6.prop.9.p1668

Let AB be the given straight line; thus it is required to cut off from AB a prescribed part.

6.prop.9.p1669

Let the third part be that prescribed.

6.prop.9.p1670

Let a straight line AC be drawn through from A containing with AB any angle; let a point D be taken at random on AC, and let DE, EC be made equal to AD. [I. 3]

6.prop.9.p1671

Let BC be joined, and through D let DF be drawn parallel to it. [I. 31]

6.prop.9.p1672

Then, since FD has been drawn parallel to BC, one of the sides of the triangle ABC, therefore, proportionally, as CD is to DA, so is BF to FA. [VI. 2]

6.prop.9.p1673

But CD is double of DA; therefore BF is also double of FA; therefore BA is triple of AF.

6.prop.9.p1674

Therefore from the given straight line

6.prop.9.p1674

AB

6.prop.9.p1674

the prescribed third part

6.prop.9.p1674

AF

6.prop.9.p1674

has been cut off. Q. E. F.

6.prop.9.p1674

1

6.prop.10.p1675

To cut a given uncut straight line similarly to a given cut straight line.

6.prop.10.p1676

Let AB be the given uncut straight line, and AC the straight line cut at the points D, E; and let them be so placed as to contain any angle; let CB be joined, and through D, E let DF, EG be drawn parallel to BC, and through D let DHK be drawn parallel to AB. [I. 31]

6.prop.10.p1677

Therefore each of the figures FH, HB is a parallelogram; therefore DH is equal to FG and HK to GB. [I. 34]

6.prop.10.p1678

Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC, therefore, proportionally, as CE is to ED, so is KH to HD. [VI. 2]

6.prop.10.p1679

But KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF.

6.prop.10.p1680

Again, since FD has been drawn parallel to GE, one of the sides of the triangle AGE, therefore, proportionally, as ED is to DA, so is GF to FA. [VI. 2]

6.prop.10.p1681

But it was also proved that, as CE is to ED, so is BG to GF; therefore, as CE is to ED, so is BG to GF, and, as ED is to DA, so is GF to FA.

6.prop.10.p1682

Therefore the given uncut straight line AB has been cut similarly to the given cut straight line AC. Q. E. F.

6.prop.11.p1683

To two given straight lines to find a third proportional.

6.prop.11.p1684

Let BA, AC be the two given straight lines, and let them be placed so as to contain any angle; thus it is required to find a third proportional to BA, AC.

6.prop.11.p1685

For let them be produced to the points D, E, and let BD be made equal to AC; [I. 3] let BC be joined, and through D let DE be drawn parallel to it. [I. 31]

6.prop.11.p1686

Since, then, BC has been drawn parallel to DE, one of the sides of the triangle ADE, proportionally, as AB is to BD, so is AC to CE. [VI. 2]

6.prop.11.p1687

But BD is equal to AC; therefore, as AB is to AC, so is AC to CE.

6.prop.11.p1688

Therefore to two given straight lines

6.prop.11.p1688

AB

6.prop.11.p1688

,

6.prop.11.p1688

AC

6.prop.11.p1688

a third proportional to them,

6.prop.11.p1688

CE

6.prop.11.p1688

, has been found. Q. E. F.

6.prop.11.p1688

1

6.prop.12.p1689

To three given straight lines to find a fourth proportional.

6.prop.12.p1690

Let A, B, C be the three given straight lines; thus it is required to find a fourth proportional to A, B, C.

6.prop.12.p1691

Let two straight lines DE, DF be set out containing any angle EDF; let DG be made equal to A, GE equal to B, and further DH equal to C; let GH be joined, and let EF be drawn through E parallel to it. [I. 31]

6.prop.12.p1692

Since, then, GH has been drawn parallel to EF, one of the sides of the triangle DEF, therefore, as DG is to GE, so is DH to HF. [VI. 2]

6.prop.12.p1693

But DG is equal to A, GE to B, and DH to C; therefore, as A is to B, so is C to HF.

6.prop.12.p1694

Therefore to the three given straight lines A, B, C a fourth proportional HF has been found. Q. E. F.

6.prop.13.p1695

To two given straight lines to find a mean proportional.

6.prop.13.p1696

Let AB, BC be the two given straight lines; thus it is required to find a mean proportional to AB, BC.

6.prop.13.p1697

Let them be placed in a straight line, and let the semicircle ADC be described on AC; let BD be drawn from the point B at right angles to the straight line AC, and let AD, DC be joined.

6.prop.13.p1698

Since the angle ADC is an angle in a semicircle, it is right. [III. 31]

6.prop.13.p1699

And, since, in the right-angled triangle ADC, DB has been drawn from the right angle perpendicular to the base, therefore DB is a mean proportional between the segments of the base, AB, BC. [VI. 8, Por.]

6.prop.13.p1700

Therefore to the two given straight lines AB, BC a mean proportional DB has been found. Q. E. F.

6.prop.14.p1701

In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.

6.prop.14.p1702

Let AB, BC be equal and equiangular parallelograms having the angles at B equal, and let DB, BE be placed in a straight line; therefore FB, BG are also in a straight line. [I. 14]

6.prop.14.p1703

I say that, in AB, BC, the sides about the equal angles are reciprocally proportional, that is to say, that, as DB is to BE, so is GB to BF.

6.prop.14.p1704

For let the parallelogram FE be completed.

6.prop.14.p1705

Since, then, the parallelogram AB is equal to the parallelogram BC, and FE is another area, therefore, as AB is to FE, so is BC to FE. [V. 7]

6.prop.14.p1706

But, as AB is to FE, so is DB to BE, [VI. 1] and, as BC is to FE, so is GB to BF. [id.] therefore also, as DB is to BE, so is GB to BF. [V. 11]

6.prop.14.p1707

Therefore in the parallelograms AB, BC the sides about the equal angles are reciprocally proportional.

6.prop.14.p1708

Next, let GB be to BF as DB to BE; I say that the parallelogram AB is equal to the parallelogram BC.

6.prop.14.p1709

For since, as DB is to BE, so is GB to BF, while, as DB is to BE, so is the parallelogram AB to the parallelogram FE, [VI. 1] and, as GB is to BF, so is the parallelogram BC to the parallelogram FE, [VI. 1] therefore also, as AB is to FE, so is BC to FE; [V. 11] therefore the parallelogram AB is equal to the parallelogram BC. [V. 9]

6.prop.14.p1710

Therefore etc. Q. E. D.

6.prop.15.p1711

In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal.

6.prop.15.p1712

Let ABC, ADE be equal triangles having one angle equal to one angle, namely the angle BAC to the angle DAE; I say that in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional, that is to say, that, as CA is to AD, so is EA to AB.

6.prop.15.p1713

For let them be placed so that CA is in a straight line with AD; therefore EA is also in a straight line with AB. [I. 14]

6.prop.15.p1714

Let BD be joined.

6.prop.15.p1715

Since then the triangle ABC is equal to the triangle ADE, and BAD is another area, therefore, as the triangle CAB is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 7]

6.prop.15.p1716

But, as CAB is to BAD, so is CA to AD, [VI. 1] and, as EAD is to BAD, so is EA to AB. [id.]

6.prop.15.p1717

Therefore also, as CA is to AD, so is EA to AB. [V. 11]

6.prop.15.p1718

Therefore in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional.

6.prop.15.p1719

Next, let the sides of the triangles ABC, ADE be reciprocally proportional, that is to say, let EA be to AB as CA to AD; I say that the triangle ABC is equal to the triangle ADE.

6.prop.15.p1720

For, if BD be again joined, since, as CA is to AD, so is EA to AB, while, as CA is to AD, so is the triangle ABC to the triangle BAD, and, as EA is to AB, so is the triangle EAD to the triangle BAD, [VI. 1] therefore, as the triangle ABC is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 11]

6.prop.15.p1721

Therefore each of the triangles ABC, EAD has the same ratio to BAD.

6.prop.15.p1722

Therefore the triangle ABC is equal to the triangle EAD. [V. 9]

6.prop.15.p1723

Therefore etc. Q. E. D.

6.prop.16.p1724

If four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means; and, if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines will be proportional.

6.prop.16.p1725

Let the four straight lines AB, CD, E, F be proportional, so that, as AB is to CD, so is E to F; I say that the rectangle contained by AB, F is equal to the rectangle contained by CD, E.

6.prop.16.p1726

Let AG, CH be drawn from the points A, C at right angles to the straight lines AB, CD, and let AG be made equal to F, and CH equal to E.

6.prop.16.p1727

Let the parallelograms BG, DH be completed.

6.prop.16.p1728

Then since, as AB is to CD, so is E to F, while E is equal to CH, and F to AG, therefore, as AB is to CD, so is CH to AG.

6.prop.16.p1729

Therefore in the parallelograms BG, DH the sides about the equal angles are reciprocally proportional.

6.prop.16.p1730

But those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal; [VI. 14] therefore the parallelogram BG is equal to the parallelogram DH.

6.prop.16.p1731

And BG is the rectangle AB, F, for AG is equal to F; and DH is the rectangle CD, E, for E is equal to CH; therefore the rectangle contained by AB, F is equal to the rectangle contained by CD, E.

6.prop.16.p1732

Next, let the rectangle contained by AB, F be equal to the rectangle contained by CD, E; I say that the four straight lines will be proportional, so that, as AB is to CD, so is E to F.

6.prop.16.p1733

For, with the same construction, since the rectangle AB, F is equal to the rectangle CD, E, and the rectangle AB, F is BG, for AG is equal to F, and the rectangle CD, E is DH, for CH is equal to E, therefore BG is equal to DH.

6.prop.16.p1734

And they are equiangular

6.prop.16.p1735

But in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. [VI. 14]

6.prop.16.p1736

Therefore, as AB is to CD, so is CH to AG.

6.prop.16.p1737

But CH is equal to E, and AG to F; therefore, as AB is to CD, so is E to F.

6.prop.16.p1738

Therefore etc. Q. E. D.

6.prop.17.p1739

If three straight lines be proportional, the rectangle contained by the extremes is equal to the square on the mean; and, if the rectangle contained by the extremes be equal to the square on the mean, the three straight lines will be proportional.

6.prop.17.p1740

Let the three straight lines A, B, C be proportional, so that, as A is to B, so is B to C; I say that the rectangle contained by A, C is equal to the square on B.

6.prop.17.p1741

Let D be made equal to B.

6.prop.17.p1742

Then, since, as A is to B, so is B to C, and B is equal to D, therefore, as A is to B, so is D to C.

6.prop.17.p1743

But, if four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means. [VI. 16]

6.prop.17.p1744

Therefore the rectangle A, C is equal to the rectangle B, D.

6.prop.17.p1745

But the rectangle B, D is the square on B, for B is equal to D; therefore the rectangle contained by A, C is equal to the square on B.

6.prop.17.p1746

Next, let the rectangle A, C be equal to the square on B; I say that, as A is to B, so is B to C.

6.prop.17.p1747

For, with the same construction, since the rectangle A, C is equal to the square on B, while the square on B is the rectangle B, D, for B is equal to D, therefore the rectangle A, C is equal to the rectangle B, D.

6.prop.17.p1748

But, if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportional. [VI. 16]

6.prop.17.p1749

Therefore, as A is to B, so is D to C.

6.prop.17.p1750

But B is equal to D; therefore, as A is to B, so is B to C.

6.prop.17.p1751

Therefore etc. Q. E. D.

6.prop.18.p1752

On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure.

6.prop.18.p1753

Let AB be the given straight line and CE the given rectilineal figure; thus it is required to describe on the straight line AB a rectilineal figure similar and similarly situated to the rectilineal figure CE.

6.prop.18.p1754

Let DF be joined, and on the straight line AB, and at the points A, B on it, let the angle GAB be constructed equal to the angle at C, and the angle ABG equal to the angle CDF. [I. 23]

6.prop.18.p1755

Therefore the remaining angle CFD is equal to the angle AGB; [I. 32] therefore the triangle FCD is equiangular with the triangle GAB.

6.prop.18.p1756

Therefore, proportionally, as FD is to GB, so is FC to GA, and CD to AB.

6.prop.18.p1757

Again, on the straight line BG, and at the points B, G on it, let the angle BGH be constructed equal to the angle DFE, and the angle GBH equal to the angle FDE. [I. 23]

6.prop.18.p1758

Therefore the remaining angle at E is equal to the remaining angle at H; [I. 32] therefore the triangle FDE is equiangular with the triangle GBH; therefore, proportionally, as FD is to GB, so is FE to GH, and ED to HB. [VI. 4]

6.prop.18.p1759

But it was also proved that, as FD is to GB, so is FC to GA, and CD to AB; therefore also, as FC is to AG, so is CD to AB, and FE to GH, and further ED to HB.

6.prop.18.p1760

And, since the angle CFD is equal to the angle AGB, and the angle DFE to the angle BGH, therefore the whole angle CFE is equal to the whole angle AGH.

6.prop.18.p1761

For the same reason the angle CDE is also equal to the angle ABH.

6.prop.18.p1762

And the angle at C is also equal to the angle at A, and the angle at E to the angle at H.

6.prop.18.p1763

Therefore AH is equiangular with CE; and they have the sides about their equal angles proportional; therefore the rectilineal figure AH is similar to the rectilineal figure CE. [VI. Def. 1]

6.prop.18.p1764

Therefore on the given straight line AB the rectilineal figure AH has been described similar and similarly situated to the given rectilineal figure CE. Q. E. F.

6.prop.19.p1765

Similar triangles are to one another in the duplicate ratio of the corresponding sides.

6.prop.19.p1766

Let ABC, DEF be similar triangles having the angle at B equal to the angle at E, and such that, as AB is to BC, so is DE to EF, so that BC corresponds to EF; [V. Def. 11] I say that the triangle ABC has to the triangle DEF a ratio duplicate of that which BC has to EF.

6.prop.19.p1767

For let a third proportional BG be taken to BC, EF, so that, as BC is to EF, so is EF to BG; [VI. 11] and let AG be joined.

6.prop.19.p1768

Since then, as AB is to BC, so is DE to EF, therefore, alternately, as AB is to DE, so is BC to EF. [V. 16]

6.prop.19.p1769

But, as BC is to EF, so is EF to BG; therefore also, as AB is to DE, so is EF to BG. [V. 11]

6.prop.19.p1770

Therefore in the triangles ABG, DEF the sides about the equal angles are reciprocally proportional.

6.prop.19.p1771

But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal; [VI. 15] therefore the triangle ABG is equal to the triangle DEF.

6.prop.19.p1772

Now since, as BC is to EF, so is EF to BG, and, if three straight lines be proportional, the first has to the third a ratio duplicate of that which it has to the second, [V. Def. 9] therefore BC has to BG a ratio duplicate of that which CB has to EF.

6.prop.19.p1773

But, as CB is to BG, so is the triangle ABC to the triangle ABG; [VI. 1] therefore the triangle ABC also has to the triangle ABG a ratio duplicate of that which BC has to EF.

6.prop.19.p1774

But the triangle ABG is equal to the triangle DEF; therefore the triangle ABC also has to the triangle DEF a ratio duplicate of that which BC has to EF.

6.prop.19.p1775

Therefore etc.

6.prop.19.p1776

Porism. From this it is manifest that, if three straight lines be proportional, then, as the first is to the third, so is the figure described on the first to that which is similar and similarly described on the second. Q. E. D.

6.prop.20.p1777

Similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side.

6.prop.20.p1778

Let ABCDE, FGHKL be similar polygons, and let AB correspond to FG; I say that the polygons ABCDE, FGHKL are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which AB has to FG.

6.prop.20.p1779

Let BE, EC, GL, LH be joined.

6.prop.20.p1780

Now, since the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL; and, as BA is to AE, so is GF to FL. [VI. Def. 1]

6.prop.20.p1781

Since then ABE, FGL are two triangles having one angle equal to one angle and the sides about the equal angles proportional, therefore the triangle ABE is equiangular with the triangle FGL; [VI. 6] so that it is also similar; [VI. 4 and Def. 1] therefore the angle ABE is equal to the angle FGL.

6.prop.20.p1782

But the whole angle ABC is also equal to the whole angle FGH because of the similarity of the polygons; therefore the remaining angle EBC is equal to the angle LGH.

6.prop.20.p1783

And, since, because of the similarity of the triangles ABE, FGL, as EB is to BA, so is LG to GF, and moreover also, because of the similarity of the polygons, as AB is to BC, so is FG to GH, therefore, ex aequali, as EB is to BC, so is LG to GH; [V. 22] that is, the sides about the equal angles EBC, LGH are proportional; therefore the triangle EBC is equiangular with the triangle LGH, [VI. 6] so that the triangle EBC is also similar to the triangle LGH. [VI. 4 and Def. 1]

6.prop.20.p1784

For the same reason the triangle ECD is also similar to the triangle LHK.

6.prop.20.p1785

Therefore the similar polygons ABCDE, FGHKL have been divided into similar triangles, and into triangles equal in multitude.

6.prop.20.p1786

I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and ABE, EBC, ECD are antecedents, while FGL, LGH, LHK are their consequents, and that the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which the corresponding side has to the corresponding side, that is AB to FG.

6.prop.20.p1787

For let AC, FH be joined.

6.prop.20.p1788

Then since, because of the similarity of the polygons, the angle ABC is equal to the angle FGH, and, as AB is to BC, so is FG to GH, the triangle ABC is equiangular with the triangle FGH; [VI. 6] therefore the angle BAC is equal to the angle GFH, and the angle BCA to the angle GHF.

6.prop.20.p1789

And, since the angle BAM is equal to the angle GFN, and the angle ABM is also equal to the angle FGN, therefore the remaining angle AMB is also equal to the remaining angle FNG; [I. 32] therefore the triangle ABM is equiangular with the triangle FGN.

6.prop.20.p1790

Similarly we can prove that the triangle BMC is also equiangular with the triangle GNH.

6.prop.20.p1791

Therefore, proportionally, as AM is to MB, so is FN to NG, and, as BM is to MC, so is GN to NH; so that, in addition, ex aequali, as AM is to MC, so is FN to NH.

6.prop.20.p1792

But, as AM is to MC, so is the triangle ABM to MBC, and AME to EMC; for they are to one another as their bases. [VI. 1]

6.prop.20.p1793

Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12] therefore, as the triangle AMB is to BMC, so is ABE to CBE.

6.prop.20.p1794

But, as AMB is to BMC, so is AM to MC; therefore also, as AM is to MC, so is the triangle ABE to the triangle EBC.

6.prop.20.p1795

For the same reason also, as FN is to NH, so is the triangle FGL to the triangle GLH.

6.prop.20.p1796

And, as AM is to MC, so is FN to NH; therefore also, as the triangle ABE is to the triangle BEC, so is the triangle FGL to the triangle GLH; and, alternately, as the triangle ABE is to the triangle FGL, so is the triangle BEC to the triangle GLH.

6.prop.20.p1797

Similarly we can prove, if BD, GK be joined, that, as the triangle BEC is to the triangle LGH, so also is the triangle ECD to the triangle LHK.

6.prop.20.p1798

And since, as the triangle ABE is to the triangle FGL, so is EBC to LGH, and further ECD to LHK, therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [V. 12 therefore, as the triangle ABE is to the triangle FGL, so is the polygon ABCDE to the polygon FGHKL.

6.prop.20.p1799

But the triangle ABE has to the triangle FGL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG; for similar triangles are in the duplicate ratio of the corresponding sides. [VI. 19]

6.prop.20.p1800

Therefore the polygon ABCDE also has to the polygon FGHKL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG.

6.prop.20.p1801

Therefore etc.

6.prop.20.p1802

Porism. Similarly also it can be proved in the case of quadrilaterals that they are in the duplicate ratio of the corresponding sides. And it was also proved in the case of triangles; therefore also, generally, similar rectilineal figures are to one another in the duplicate ratio of the corresponding sides. Q. E. D.

6.prop.21.p1803

Figures which are similar to the same rectilineal figure are also similar to one another.

6.prop.21.p1804

For let each of the rectilineal figures A, B be similar to C; I say that A is also similar to B.

6.prop.21.p1805

For, since A is similar to C, it is equiangular with it and has the sides about the equal angles proportional. [VI. Def. 1]

6.prop.21.p1806

Again, since B is similar to C, it is equiangular with it and has the sides about the equal angles proportional.

6.prop.21.p1807

Therefore each of the figures A, B is equiangular with C and with C has the sides about the equal angles proportional; therefore A is similar to B. Q. E. D.

6.prop.22.p1808

If four straight lines be proportional, the rectilineal figures similar and similarly described upon them will also be proportional; and, if the rectilineal figures similar and similarly described upon them be proportional, the straight lines will themselves also be proportional.

6.prop.22.p1809

Let the four straight lines AB, CD, EF, GH be proportional, so that, as AB is to CD, so is EF to GH, and let there be described on AB, CD the similar and similarly situated rectilineal figures KAB, LCD, and on EF, GH the similar and similarly situated rectilineal figures MF, NH; I say that, as KAB is to LCD, so is MF to NH.

6.prop.22.p1810

For let there be taken a third proportional O to AB, CD, and a third proportional P to EF, GH. [VI. 11]

6.prop.22.p1811

Then since, as AB is to CD, so is EF to GH, and, as CD is to O, so is GH to P, therefore, ex aequali, as AB is to O, so is EF to P. [V. 22]

6.prop.22.p1812

But, as AB is to O, so is KAB to LCD, [VI. 19, Por.] and, as EF is to P, so is MF to NH; therefore also, as KAB is to LCD, so is MF to NH. [V. 11]

6.prop.22.p1813

Next, let MF be to NH as KAB is to LCD; I say also that, as AB is to CD, so is EF to GH.

6.prop.22.p1814

For, if EF is not to GH as AB to CD, let EF be to QR as AB to CD, [VI. 12] and on QR let the rectilineal figure SR be described similar and similarly situated to either of the two MF, NH. [VI. 18]

6.prop.22.p1815

Since then, as AB is to CD, so is EF to QR, and there have been described on AB, CD the similar and similarly situated figures KAB, LCD, and on EF, QR the similar and similarly situated figures MF, SR, therefore, as KAB is to LCD, so is MF to SR.

6.prop.22.p1816

But also, by hypothesis, as KAB is to LCD, so is MF to NH; therefore also, as MF is to SR, so is MF to NH. [V. 11]

6.prop.22.p1817

Therefore MF has the same ratio to each of the figures NH, SR; therefore NH is equal to SR. [V. 9]

6.prop.22.p1818

But it is also similar and similarly situated to it; therefore GH is equal to QR.

6.prop.22.p1819

And, since, as AB is to CD, so is EF to QR, while QR is equal to GH, therefore, as AB is to CD, so is EF to GH.

6.prop.22.p1820

Therefore etc. Q. E. D.

6.prop.23.p1821

Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides.

6.prop.23.p1822

Let AC, CF be equiangular parallelograms having the angle BCD equal to the angle ECG; I say that the parallelogram AC has to the parallelogram CF the ratio compounded of the ratios of the sides.

6.prop.23.p1823

For let them be placed so that BC is in a straight line with CG; therefore DC is also in a straight line with CE.

6.prop.23.p1824

Let the parallelogram DG be completed; let a straight line K be set out, and let it be contrived that, as BC is to CG, so is K to L, and, as DC is to CE, so is L to M. [VI. 12]

6.prop.23.p1825

Then the ratios of K to L and of L to M are the same as the ratios of the sides, namely of BC to CG and of DC to CE.

6.prop.23.p1826

But the ratio of K to M is compounded of the ratio of K to L and of that of L to M; so that K has also to M the ratio compounded of the ratios of the sides.

6.prop.23.p1827

Now since, as BC is to CG, so is the parallelogram AC to the parallelogram CH, [VI. 1] while, as BC is to CG, so is K to L, therefore also, as K is to L, so is AC to CH. [V. 11]

6.prop.23.p1828

Again, since, as DC is to CE, so is the parallelogram CH to CF, [VI. 1] while, as DC is to CE, so is L to M, therefore also, as L is to M, so is the parallelogram CH to the parallelogram CF. [V. 11]

6.prop.23.p1829

Since then it was proved that, as K is to L, so is the parallelogram AC to the parallelogram CH, and, as L is to M, so is the parallelogram CH to the parallelogram CF, therefore, ex aequali, as K is to M, so is AC to the parallelogram CF.

6.prop.23.p1830

But K has to M the ratio compounded of the ratios of the sides; therefore AC also has to CF the ratio compounded of the ratios of the sides.

6.prop.23.p1831

Therefore etc. Q. E. D.

6.prop.23.p1831

1

6.prop.23.p1831

2

6.prop.24.p1832

In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another.

6.prop.24.p1833

Let ABCD be a parallelogram, and AC its diameter, and let EG, HK be parallelograms about AC; I say that each of the parallelograms EG, HK is similar both to the whole ABCD and to the other.

6.prop.24.p1834

For, since EF has been drawn parallel to BC, one of the sides of the triangle ABC, proportionally, as BE is to EA, so is CF to FA. [VI. 2]

6.prop.24.p1835

Again, since FG has been drawn parallel to CD, one of the sides of the triangle ACD, proportionally, as CF is to FA, so is DG to GA. [VI. 2]

6.prop.24.p1836

But it was proved that, as CF is to FA, so also is BE to EA; therefore also, as BE is to EA, so is DG to GA, and therefore, componendo, as BA is to AE, so is DA to AG, [V. 18] and, alternately, as BA is to AD, so is EA to AG. [V. 16]

6.prop.24.p1837

Therefore in the parallelograms ABCD, EG, the sides about the common angle BAD are proportional.

6.prop.24.p1838

And, since GF is parallel to DC, the angle AFG is equal to the angle DCA; and the angle DAC is common to the two triangles ADC, AGF; therefore the triangle ADC is equiangular with the triangle AGF.

6.prop.24.p1839

For the same reason the triangle ACB is also equiangular with the triangle AFE, and the whole parallelogram ABCD is equiangular with the parallelogram EG.

6.prop.24.p1840

Therefore, proportionally, as AD is to DC, so is AG to GF, as DC is to CA, so is GF to FA, as AC is to CB, so is AF to FE, and further, as CB is to BA, so is FE to EA.

6.prop.24.p1841

And, since it was proved that, as DC is to CA, so is GF to FA, and, as AC is to CB, so is AF to FE, therefore, ex aequali, as DC is to CB, so is GF to FE. [V. 22]

6.prop.24.p1842

Therefore in the parallelograms ABCD, EG the sides about the equal angles are proportional; therefore the parallelogram ABCD is similar to the parallelogram EG. [VI. Def. 1]

6.prop.24.p1843

For the same reason the parallelogram ABCD is also similar to the parallelogram KH; therefore each of the parallelograms EG, HK is similar to ABCD.

6.prop.24.p1844

But figures similar to the same rectilineal figure are also similar to one another; [VI. 21] therefore the parallelogram EG is also similar to the parallelogram HK.

6.prop.24.p1845

Therefore etc. Q. E. D.

6.prop.25.p1846

To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure.

6.prop.25.p1847

Let ABC be the given rectilineal figure to which the figure to be constructed must be similar, and D that to which it must be equal; thus it is required to construct one and the same figure similar to ABC and equal to D.

6.prop.25.p1848

Let there be applied to BC the parallelogram BE equal to the triangle ABC [I. 44], and to CE the parallelogram CM equal to D in the angle FCE which is equal to the angle CBL. [I. 45]

6.prop.25.p1849

Therefore BC is in a straight line with CF, and LE with EM.

6.prop.25.p1850

Now let GH be taken a mean proportional to BC, CF [VI. 13], and on GH let KGH be described similar and similarly situated to ABC. [VI. 18]

6.prop.25.p1851

Then, since, as BC is to GH, so is GH to CF, and, if three straight lines be proportional, as the first is to the third, so is the figure on the first to the similar and similarly situated figure described on the second, [VI. 19, Por.] therefore, as BC is to CF, so is the triangle ABC to the triangle KGH.

6.prop.25.p1852

But, as BC is to CF, so also is the parallelogram BE to the parallelogram EF. [VI. 1]

6.prop.25.p1853

Therefore also, as the triangle ABC is to the triangle KGH, so is the parallelogram BE to the parallelogram EF; therefore, alternately, as the triangle ABC is to the parallelogram BE, so is the triangle KGH to the parallelogram EF. [V. 16]

6.prop.25.p1854

But the triangle ABC is equal to the parallelogram BE; therefore the triangle KGH is also equal to the parallelogram EF.

6.prop.25.p1855

But the parallelogram EF is equal to D; therefore KGH is also equal to D.

6.prop.25.p1856

And KGH is also similar to ABC.

6.prop.25.p1857

Therefore one and the same figure

6.prop.25.p1857

KGH

6.prop.25.p1857

has been constructed similar to the given rectilineal figure

6.prop.25.p1857

ABC

6.prop.25.p1857

and equal to the other given figure

6.prop.25.p1857

D

6.prop.25.p1857

. Q. E. D.

6.prop.25.p1857

1

6.prop.26.p1858

If from a parallelogram there be taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, it is about the same diameter with the whole

6.prop.26.p1859

For from the parallelogram ABCD let there be taken away the parallelogram AF similar and similarly situated to ABCD, and having the angle DAB common with it; I say that ABCD is about the same diameter with AF.

6.prop.26.p1860

For suppose it is not, but, if possible, let AHC be the diameter lt of ABCD gt, let GF be produced and carried through to H, and let HK be drawn through H parallel to either of the straight lines AD, BC. [I. 31]

6.prop.26.p1861

Since, then, ABCD is about the same diameter with KG, therefore, as DA is to AB, so is GA to AK. [VI. 24]

6.prop.26.p1862

But also, because of the similarity of ABCD, EG, as DA is to AB, so is GA to AE; therefore also, as GA is to AK, so is GA to AE. [V. 11]

6.prop.26.p1863

Therefore GA has the same ratio to each of the straight lines AK, AE.

6.prop.26.p1864

Therefore AE is equal to AK [V. 9], the less to the greater : which is impossible.

6.prop.26.p1865

Therefore ABCD cannot but be about the same diameter with AF; therefore the parallelogram ABCD is about the same diameter with the parallelogram AF.

6.prop.26.p1866

Therefore etc. Q. E. D.

6.prop.27.p1867

Of all the parallelograms applied to the same straight line and deficient by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the defect.

6.prop.27.p1868

Let AB be a straight line and let it be bisected at C; let there be applied to the straight line AB the parallelogram AD deficient by the parallelogrammic figure DB described on the half of AB, that is, CB; I say that, of all the parallelograms applied to AB and deficient by parallelogrammic figures similar and similarly situated to DB, AD is greatest.

6.prop.27.p1869

For let there be applied to the straight line AB the parallelogram AF deficient by the parallelogrammic figure FB similar and similarly situated to DB; I say that AD is greater than AF.

6.prop.27.p1870

For, since the parallelogram DB is similar to the parallelogram FB, they are about the same diameter. [VI. 26]

6.prop.27.p1871

Let their diameter DB be drawn, and let the figure be described.

6.prop.27.p1872

Then, since CF is equal to FE, [I. 43] and FB is common, therefore the whole CH is equal to the whole KE.

6.prop.27.p1873

But CH is equal to CG, since AC is also equal to CB. [I. 36]

6.prop.27.p1874

Therefore GC is also equal to EK.

6.prop.27.p1875

Let CF be added to each; therefore the whole AF is equal to the gnomon LMN; so that the parallelogram DB, that is, AD, is greater than the parallelogram AF.

6.prop.27.p1876

Therefore etc.

6.prop.28.p1877

To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one : thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect.

6.prop.28.p1878

Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, not being greater than the parallelogram described on the half of AB and similar to the defect, and D the parallelogram to which the defect is required to be similar; thus it is required to apply to the given straight line AB a parallelogram equal to the given rectilineal figure C and deficient by a parallelogrammic figure which is similar to D.

6.prop.28.p1879

Let AB be bisected at the point E, and on EB let EBFG be described similar and similarly situated to D; [VI. 18] let the parallelogram AG be completed.

6.prop.28.p1880

If then AG is equal to C, that which was enjoined will have been done; for there has been applied to the given straight line AB the parallelogram AG equal to the given rectilineal figure C and deficient by a parallelogrammic figure GB which is similar to D.

6.prop.28.p1881

But, if not, let HE be greater than C.

6.prop.28.p1882

Now HE is equal to GB; therefore GB is also greater than C.

6.prop.28.p1883

Let KLMN be constructed at once equal to the excess by which GB is greater than C and similar and similarly situated to D. [VI. 25]

6.prop.28.p1884

But D is similar to GB; therefore KM is also similar to GB. [VI. 21]

6.prop.28.p1885

Let, then, KL correspond to GE, and LM to GF.

6.prop.28.p1886

Now, since GB is equal to C, KM, therefore GB is greater than KM; therefore also GE is greater than KL, and GF than LM.

6.prop.28.p1887

Let GO be made equal to KL, and GP equal to LM; and let the parallelogram OGPQ be completed; therefore it is equal and similar to KM.

6.prop.28.p1888

Therefore GQ is also similar to GB; [VI. 21] therefore GQ is about the same diameter with GB. [VI. 26]

6.prop.28.p1889

Let GQB be their diameter, and let the figure be described.

6.prop.28.p1890

Then, since BG is equal to C, KM, and in them GQ is equal to KM, therefore the remainder, the gnomon UWV, is equal to the remainder C.

6.prop.28.p1891

And, since PR is equal to OS, let QB be added to each; therefore the whole PB is equal to the whole OB.

6.prop.28.p1892

But OB is equal to TE, since the side AE is also equal to the side EB; [I. 36] therefore TE is also equal to PB.

6.prop.28.p1893

Let OS be added to each; therefore the whole TS is equal to the whole, the gnomon VWU.

6.prop.28.p1894

But the gnomon VWU was proved equal to C; therefore TS is also equal to C.

6.prop.28.p1895

Therefore to the given straight line AB there has been applied the parallelogram ST equal to the given rectilineal figure C and deficient by a parallelogrammic figure QB which is similar to D. Q. E. F.

6.prop.29.p1896

To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogrammic figure similar to a given one.

6.prop.29.p1897

Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, and D that to which the excess is required to be similar; thus it is required to apply to the straight line AB a parallelogram equal to the rectilineal figure C and exceeding by a parallelogrammic figure similar to D.

6.prop.29.p1898

Let AB be bisected at E; let there be described on EB the parallelogram BF similar and similarly situated to D; and let GH be constructed at once equal to the sum of BF, C and similar and similarly situated to D. [VI. 25]

6.prop.29.p1899

Let KH correspond to FL and KG to FE.

6.prop.29.p1900

Now, since GH is greater than FB, therefore KH is also greater than FL, and KG than FE.

6.prop.29.p1901

Let FL, FE be produced, let FLM be equal to KH, and FEN to KG, and let MN be completed; therefore MN is both equal and similar to GH.

6.prop.29.p1902

But GH is similar to EL; therefore MN is also similar to EL; [VI. 21] therefore EL is about the same diameter with MN. [VI. 26]

6.prop.29.p1903

Let their diameter FO be drawn, and let the figure be described.

6.prop.29.p1904

Since GH is equal to EL, C, while GH is equal to MN, therefore MN is also equal to EL, C.

6.prop.29.p1905

Let EL be subtracted from each; therefore the remainder, the gnomon XWV, is equal to C.

6.prop.29.p1906

Now, since AE is equal to EB, AN is also equal to NB [I. 36], that is, to LP [I. 43].

6.prop.29.p1907

Let EO be added to each; therefore the whole AO is equal to the gnomon VWX.

6.prop.29.p1908

But the gnomon VWX is equal to C; therefore AO is also equal to C.

6.prop.29.p1909

Therefore to the given straight line AB there has been applied the parallelogram AO equal to the given rectilineal figure C and exceeding by a parallelogrammic figure QP which is similar to D, since PQ is also similar to EL [VI. 24]. Q. E. F.

6.prop.30.p1910

To cut a given finite straight line in extreme and mean ratio.

6.prop.30.p1911

Let AB be the given finite straight line; thus it is required to cut AB in extreme and mean ratio.

6.prop.30.p1912

On AB let the square BC be described; and let there be applied to AC the parallelogram CD equal to BC and exceeding by the figure AD similar to BC. [VI. 29]

6.prop.30.p1913

Now BC is a square; therefore AD is also a square.

6.prop.30.p1914

And, since BC is equal to CD, let CE be subtracted from each; therefore the remainder BF is equal to the remainder AD.

6.prop.30.p1915

But it is also equiangular with it; therefore in BF, AD the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, as FE is to ED, so is AE to EB.

6.prop.30.p1916

But FE is equal to AB, and ED to AE.

6.prop.30.p1917

Therefore, as BA is to AE, so is AE to EB.

6.prop.30.p1918

And AB is greater than AE; therefore AE is also greater than EB.

6.prop.30.p1919

Therefore the straight line AB has been cut in extreme and mean ratio at E, and the greater segment of it is AE. Q. E. F.

6.prop.31.p1920

In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.

6.prop.31.p1921

Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA, AC.

6.prop.31.p1922

Let AD be drawn perpendicular.

6.prop.31.p1923

Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another. [VI. 8]

6.prop.31.p1924

And, since ABC is similar to ABD, therefore, as CB is to BA, so is AB to BD. [VI. Def. 1]

6.prop.31.p1925

And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second. [VI. 19, Por.]

6.prop.31.p1926

Therefore, as CB is to BD, so is the figure on CB to the similar and similarly described figure on BA.

6.prop.31.p1927

For the same reason also, as BC is to CD, so is the figure on BC to that on CA; so that, in addition, as BC is to BD, DC, so is the figure on BC to the similar and similarly described figures on BA, AC.

6.prop.31.p1928

But BC is equal to BD, DC; therefore the figure on BC is also equal to the similar and similarly described figures on BA, AC.

6.prop.31.p1929

Therefore etc. Q. E. D.

6.prop.32.p1930

If two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line.

6.prop.32.p1931

Let ABC, DCE be two triangles having the two sides BA, AC proportional to the two sides DC, DE, so that, as AB is to AC, so is DC to DE, and AB parallel to DC, and AC to DE; I say that BC is in a straight line with CE.

6.prop.32.p1932

For, since AB is parallel to DC, and the straight line AC has fallen upon them, the alternate angles BAC, ACD are equal to one another. [I. 29]

6.prop.32.p1933

For the same reason the angle CDE is also equal to the angle ACD; so that the angle BAC is equal to the angle CDE.

6.prop.32.p1934

And, since ABC, DCE are two triangles having one angle, the angle at A, equal to one angle, the angle at D, and the sides about the equal angles proportional, so that, as BA is to AC, so is CD to DE, therefore the triangle ABC is equiangular with the triangle DCE; [VI. 6] therefore the angle ABC is equal to the angle DCE.

6.prop.32.p1935

But the angle ACD was also proved equal to the angle BAC; therefore the whole angle ACE is equal to the two angles ABC, BAC.

6.prop.32.p1936

Let the angle ACB be added to each; therefore the angles ACE, ACB are equal to the angles BAC, ACB, CBA.

6.prop.32.p1937

But the angles BAC, ABC, ACB are equal to two right angles; [I. 32] therefore the angles ACE, ACB are also equal to two right angles.

6.prop.32.p1938

Therefore with a straight line AC, and at the point C on it, the two straight lines BC, CE not lying on the same side make the adjacent angles ACE, ACB equal to two right angles; therefore BC is in a straight line with CE. [I. 14]

6.prop.32.p1939

Therefore etc. Q. E. D.

6.prop.33.p1940

In equal circles angles have the same ratio as the circumferences on which they stand, whether they stand at the centres or at the circumferences.

6.prop.33.p1941

Let ABC, DEF be equal circles, and let the angles BGC, EHF be angles at their centres G, H, and the angles BAC, EDF angles at the circumferences; I say that, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

6.prop.33.p1942

For let any number of consecutive circumferences CK, KL be made equal to the circumference BC, and any number of consecutive circumferences FM, MN equal to the circumference EF; and let GK, GL, HM, HN be joined.

6.prop.33.p1943

Then, since the circumferences BC, CK, KL are equal to one another, the angles BGC, CGK, KGL are also equal to one another; [III. 27] therefore, whatever multiple the circumference BL is of BC, that multiple also is the angle BGL of the angle BGC.

6.prop.33.p1944

For the same reason also, whatever multiple the circumference NE is of EF, that multiple also is the angle NHE of the angle EHF.

6.prop.33.p1945

If then the circumference BL is equal to the circumference EN, the angle BGL is also equal to the angle EHN; [III. 27] if the circumference BL is greater than the circumference EN, the angle BGL is also greater than the angle EHN; and, if less, less.

6.prop.33.p1946

There being then four magnitudes, two circumferences BC, EF, and two angles BGC, EHF, there have been taken, of the circumference BC and the angle BGC equimultiples, namely the circumference BL and the angle BGL, and of the circumference EF and the angle EHF equimultiples, namely the circumference EN and the angle EHN.

6.prop.33.p1947

And it has been proved that, if the circumference BL is in excess of the circumference EN, the angle BGL is also in excess of the angle EHN; if equal, equal; and if less, less.

6.prop.33.p1948

Therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF. [V. Def. 5]

6.prop.33.p1949

But, as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF; for they are doubles respectively.

6.prop.33.p1950

Therefore also, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

6.prop.33.p1951

Therefore etc. Q. E. D.

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