Elementa

Euclid

Elementa

Euclid · Elem 4 · perseus-eng2 · More by Euclid

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Original / primary: Perseus Eng2

4.def.1.p1072

A rectilineal figure is said to be inscribed in a rectilineal figure when the respective angles of the inscribed figure lie on the respective sides of that in which it is inscribed.

4.def.2.p1073

Similarly a figure is said to be circumscribed about a figure when the respective sides of the circumscribed figure pass through the respective angles of that about which it is circumscribed.

4.def.3.p1074

A rectilineal figure is said to be inscribed in a circle when each angle of the inscribed figure lies on the circumference of the circle.

4.def.4.p1075

A rectilineal figure is said to be circumscribed about a circle, when each side of the circumscribed figure touches the circumference of the circle.

4.def.5.p1076

Similarly a circle is said to be inscribed in a figure when the circumference of the circle touches each side of the figure in which it is inscribed.

4.def.6.p1077

A circle is said to be circumscribed about a figure when the circumference of the circle passes through each angle of the figure about which it is circumscribed.

4.def.7.p1078

A straight line is said to be fitted into a circle when its extremities are on the circumference of the circle.

4.prop.1.p1079

Into a given circle to fit a straight line equal to a given straight line which is not greater than the diameter of the circle.

4.prop.1.p1080

Let ABC be the given circle, and D the given straight line not greater than the diameter of the circle; thus it is required to fit into the circle ABC a straight line equal to the straight line D.

4.prop.1.p1081

Let a diameter BC of the circle ABC be drawn.

4.prop.1.p1082

Then, if BC is equal to D, that which was enjoined will have been done; for BC has been fitted into the circle ABC equal to the straight line D.

4.prop.1.p1083

But, if BC is greater than D, let CE be made equal to D, and with centre C and distance CE let the circle EAF be described; let CA be joined.

4.prop.1.p1084

Then, since the point C is the centre of the circle EAF, CA is equal to CE.

4.prop.1.p1085

But CE is equal to D; therefore D is also equal to CA.

4.prop.1.p1086

Therefore into the given circle ABC there has been fitted CA equal to the given straight line D.

4.prop.2.p1087

In a given circle to inscribe a triangle equiangular with a given triangle.

4.prop.2.p1088

Let ABC be the given circle, and DEF the given triangle; thus it is required to inscribe in the circle ABC a triangle equiangular with the triangle DEF.

4.prop.2.p1089

Let GH be drawn touching the circle ABC at A [III. 16, Por.]; on the straight line AH, and at the point A on it, let the angle HAC be constructed equal to the angle DEF, and on the straight line AG, and at the point A on it, let the angle GAB be constructed equal to the angle DFE; [I. 23] let BC be joined.

4.prop.2.p1090

Then, since a straight line AH touches the circle ABC, and from the point of contact at A the straight line AC is drawn across in the circle, therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle. [III. 32]

4.prop.2.p1091

But the angle HAC is equal to the angle DEF; therefore the angle ABC is also equal to the angle DEF.

4.prop.2.p1092

For the same reason the angle ACB is also equal to the angle DFE; therefore the remaining angle BAC is also equal to the remaining angle EDF. [I. 32]

4.prop.2.p1093

Therefore in the given circle there has been inscribed a triangle equiangular with the given triangle. Q. E. F.

4.prop.3.p1094

About a given circle to circumscribe a triangle equiangular with a given triangle.

4.prop.3.p1095

Let ABC be the given circle, and DEF the given triangle; thus it is required to circumscribe about the circle ABC a triangle equiangular with the triangle DEF.

4.prop.3.p1096

Let EF be produced in both directions to the points G, H, let the centre K of the circle ABC be taken [III. 1], and let the straight line KB be drawn across at random; on the straight line KB, and at the point K on it, let the angle BKA be constructed equal to the angle DEG, and the angle BKC equal to the angle DFH; [I. 23] and through the points A, B, C let LAM, MBN, NCL be drawn touching the circle ABC. [III. 16, Por.]

4.prop.3.p1097

Now, since LM, MN, NL touch the circle ABC at the points A, B, C, and KA, KB, KC have been joined from the centre K to the points A, B, C, therefore the angles at the points A, B, C are right. [III. 18]

4.prop.3.p1098

And, since the four angles of the quadrilateral AMBK are equal to four right angles, inasmuch as AMBK is in fact divisible into two triangles, and the angles KAM, KBM are right, therefore the remaining angles AKB, AMB are equal to two right angles.

4.prop.3.p1099

But the angles DEG, DEF are also equal to two right angles; [I. 13] therefore the angles AKB, AMB are equal to the angles DEG, DEF, of which the angle AKB is equal to the angle DEG; therefore the angle AMB which remains is equal to the angle DEF which remains.

4.prop.3.p1100

Similarly it can be proved that the angle LNB is also equal to the angle DFE; therefore the remaining angle MLN is equal to the angle EDF. [I. 32]

4.prop.3.p1101

Therefore the triangle LMN is equiangular with the triangle DEF; and it has been circumscribed about the circle ABC.

4.prop.3.p1102

Therefore about a given circle there has been circumscribed a triangle equiangular with the given triangle. Q. E. F.

4.prop.3.p1102

1

4.prop.3.p1102

2

4.prop.4.p1103

In a given triangle to inscribe a circle.

4.prop.4.p1104

Let ABC be the given triangle; thus it is required to inscribe a circle in the triangle ABC.

4.prop.4.p1105

Let the angles ABC, ACB be bisected by the straight lines BD, CD [I. 9], and let these meet one another at the point D; from D let DE, DF, DG be drawn perpendicular to the straight lines AB, BC, CA.

4.prop.4.p1106

Now, since the angle ABD is equal to the angle CBD, and the right angle BED is also equal to the right angle BFD, EBD, FBD are two triangles having two angles equal to two angles and one side equal to one side, namely that subtending one of the equal angles, which is BD common to the triangles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore DE is equal to DF.

4.prop.4.p1107

For the same reason DG is also equal to DF.

4.prop.4.p1108

Therefore the three straight lines DE, DF, DG are equal to one another; therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will pass also through the remaining points, and will touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right.

4.prop.4.p1109

For, if it cuts them, the straight line drawn at right angles to the diameter of the circle from its extremity will be found to fall within the circle : which was proved absurd; [III. 16] therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will not cut the straight lines AB, BC, CA; therefore it will touch them, and will be the circle inscribed in the triangle ABC. [IV. Def. 5]

4.prop.4.p1110

Let it be inscribed, as FGE.

4.prop.4.p1111

Therefore in the given triangle

4.prop.4.p1111

ABC

4.prop.4.p1111

the circle

4.prop.4.p1111

EFG

4.prop.4.p1111

has been inscribed. Q. E. F.

4.prop.4.p1111

1

4.prop.5.p1112

About a given triangle to circumscribe a circle.

4.prop.5.p1113

Let ABC be the given triangle; thus it is required to circumscribe a circle about the given triangle ABC.

4.prop.5.p1114

Let the straight lines AB, AC be bisected at the points D, E [I. 10], and from the points D, E let DF, EF be drawn at right angles to AB, AC; they will then meet within the triangle ABC, or on the straight line BC, or outside BC.

4.prop.5.p1115

First let them meet within at F, and let FB, FC, FA be joined.

4.prop.5.p1116

Then, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base FB. [I. 4]

4.prop.5.p1117

Similarly we can prove that CF is also equal to AF; so that FB is also equal to FC; therefore the three straight lines FA, FB, FC are equal to one another.

4.prop.5.p1118

Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and the circle will have been circumscribed about the triangle ABC.

4.prop.5.p1119

Let it be circumscribed, as ABC.

4.prop.5.p1120

Next, let DF, EF meet on the straight line BC at F, as is the case in the second figure; and let AF be joined.

4.prop.5.p1121

Then, similarly, we shall prove that the point F is the centre of the circle circumscribed about the triangle ABC.

4.prop.5.p1122

Again, let DF, EF meet outside the triangle ABC at F, as is the case in the third figure, and let AF, BF, CF be joined.

4.prop.5.p1123

Then again, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base BF. [I. 4]

4.prop.5.p1124

Similarly we can prove that CF is also equal to AF; so that BF is also equal to FC; therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and will have been circumscribed about the triangle ABC.

4.prop.5.p1125

Therefore about the given triangle a circle has been circumscribed. Q. E. F.

4.prop.5.p1126

And it is manifest that, when the centre of the circle falls within the triangle, the angle BAC, being in a segment greater than the semicircle, is less than a right angle; when the centre falls on the straight line BC, the angle BAC, being in a semicircle, is right; and when the centre of the circle falls outside the triangle, the angle BAC, being in a segment less than the semicircle, is greater than a right angle. [III. 31]

4.prop.6.p1127

In a given circle to inscribe a square.

4.prop.6.p1128

Let ABCD be the given circle; thus it is required to inscribe a square in the circle ABCD.

4.prop.6.p1129

Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and let AB, BC, CD, DA be joined.

4.prop.6.p1130

Then, since BE is equal to ED, for E is the centre, and EA is common and at right angles, therefore the base AB is equal to the base AD. [I. 4]

4.prop.6.p1131

For the same reason each of the straight lines BC, CD is also equal to each of the straight lines AB, AD; therefore the quadrilateral ABCD is equilateral.

4.prop.6.p1132

I say next that it is also right-angled.

4.prop.6.p1133

For, since the straight line BD is a diameter of the circle ABCD, therefore BAD is a semicircle; therefore the angle BAD is right. [III. 31]

4.prop.6.p1134

For the same reason each of the angles ABC, BCD, CDA is also right; therefore the quadrilateral ABCD is right-angled.

4.prop.6.p1135

But it was also proved equilateral; therefore it is a square; [I. Def. 22] and it has been inscribed in the circle ABCD.

4.prop.6.p1136

Therefore in the given circle the square ABCD has been inscribed. Q. E. F.

4.prop.7.p1137

About a given circle to circumscribe a square.

4.prop.7.p1138

Let ABCD be the given circle; thus it is required to circumscribe a square about the circle ABCD.

4.prop.7.p1139

Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and through the points A, B, C, D let FG, GH, HK, KF be drawn touching the circle ABCD. [III. 16, Por.]

4.prop.7.p1140

Then, since FG touches the circle ABCD, and EA has been joined from the centre E to the point of contact at A, therefore the angles at A are right. [III. 18]

4.prop.7.p1141

For the same reason the angles at the points B, C, D are also right.

4.prop.7.p1142

Now, since the angle AEB is right, and the angle EBG is also right, therefore GH is parailel to AC. [I. 28]

4.prop.7.p1143

For the same reason AC is also parallel to FK, so that GH is also parallel to FK. [I. 30]

4.prop.7.p1144

Similarly we can prove that each of the straight lines GF, HK is parallel to BED.

4.prop.7.p1145

Therefore GK, GC, AK, FB, BK are parallelograms; therefore GF is equal to HK, and GH to FK. [I. 34]

4.prop.7.p1146

And, since AC is equal to BD, and AC is also equal to each of the straight lines GH, FK, while BD is equal to each of the straight lines GF, HK, [I. 34] therefore the quadrilateral FGHK is equilateral.

4.prop.7.p1147

I say next that it is also right-angled.

4.prop.7.p1148

For, since GBEA is a parallelogram, and the angle AEB is right, therefore the angle AGB is also right. [I. 34]

4.prop.7.p1149

Similarly we can prove that the angles at H, K, F are also right.

4.prop.7.p1150

Therefore FGHK is right-angled.

4.prop.7.p1151

But it was also proved equilateral; therefore it is a square; and it has been circumscribed about the circle ABCD.

4.prop.7.p1152

Therefore about the given circle a square has been circumscribed. Q. E. F.

4.prop.8.p1153

In a given square to inscribe a circle.

4.prop.8.p1154

Let ABCD be the given square; thus it is required to inscribe a circle in the given square ABCD.

4.prop.8.p1155

Let the straight lines AD, AB be bisected at the points E, F respectively [I. 10], through E let EH be drawn parallel to either AB or CD, and through F let FK be drawn parallel to either AD or BC; [I. 31] therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their opposite sides are evidently equal. [I. 34]

4.prop.8.p1156

Now, since AD is equal to AB, and AE is half of AD, and AF half of AB, therefore AE is equal to AF, so that the opposite sides are also equal; therefore FG is equal to GE.

4.prop.8.p1157

Similarly we can prove that each of the straight lines GH, GK is equal to each of the straight lines FG, GE; therefore the four straight lines GE, GF, GH, GK are equal to one another.

4.prop.8.p1158

Therefore the circle described with centre G and distance one of the straight lines GE, GF, GH, GK will pass also through the remaining points.

4.prop.8.p1159

And it will touch the straight lines AB, BC, CD, DA, because the angles at E, F, H, K are right.

4.prop.8.p1160

For, if the circle cuts AB, BC, CD, DA, the straight line drawn at right angles to the diameter of the circle from its extremity will fall within the circle : which was proved absurd; [III. 16] therefore the circle described with centre G and distance one of the straight lines GE, GF, GH, GK will not cut the straight lines AB, BC, CD, DA.

4.prop.8.p1161

Therefore it will touch them, and will have been inscribed in the square ABCD.

4.prop.8.p1162

Therefore in the given square a circle has been inscribed. Q. E. F.

4.prop.9.p1163

About a given square to circumscribe a circle.

4.prop.9.p1164

Let ABCD be the given square; thus it is required to circumscribe a circle about the square ABCD.

4.prop.9.p1165

For let AC, BD be joined, and let them cut one another at E.

4.prop.9.p1166

Then, since DA is equal to AB, and AC is common, therefore the two sides DA, AC are equal to the two sides BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8]

4.prop.9.p1167

Therefore the angle DAB is bisected by AC.

4.prop.9.p1168

Similarly we can prove that each of the angles ABC, BCD, CDA is bisected by the straight lines AC, DB.

4.prop.9.p1169

Now, since the angle DAB is equal to the angle ABC, and the angle EAB is half the angle DAB, and the angle EBA half the angle ABC, therefore the angle EAB is also equal to the angle EBA; so that the side EA is also equal to EB. [I. 6]

4.prop.9.p1170

Similarly we can prove that each of the straight lines EA, EB is equal to each of the straight lines EC, ED.

4.prop.9.p1171

Therefore the four straight lines EA, EB, EC, ED are equal to one another.

4.prop.9.p1172

Therefore the circle described with centre E and distance one of the straight lines EA, EB, EC, ED will pass also through the remaining points; and it will have been circumscribed about the square ABCD.

4.prop.9.p1173

Let it be circumscribed, as ABCD.

4.prop.9.p1174

Therefore about the given square a circle has been circumscribed. Q. E. F.

4.prop.10.p1175

To construct an isosceles triangle having each of the angles at the base double of the remaining one.

4.prop.10.p1176

Let any straight line AB be set out, and let it be cut at the point C so that the rectangle contained by AB, BC is equal to the square on CA; [II. 11] with centre A and distance AB let the circle BDE be described, and let there be fitted in the circle BDE the straight line BD equal to the straight line AC which is not greater than the diameter of the circle BDE. [IV. 1]

4.prop.10.p1177

Let AD, DC be joined, and let the circle ACD be circumscribed about the triangle ACD. [IV. 5]

4.prop.10.p1178

Then, since the rectangle AB, BC is equal to the square on AC, and AC is equal to BD, therefore the rectangle AB, BC is equal to the square on BD.

4.prop.10.p1179

And, since a point B has been taken outside the circle ACD, and from B the two straight lines BA, BD have fallen on the circle ACD, and one of them cuts it, while the other falls on it, and the rectangle AB, BC is equal to the square on BD, therefore BD touches the circle ACD. [III. 37]

4.prop.10.p1180

Since, then, BD touches it, and DC is drawn across from the point of contact at D, therefore the angle BDC is equal to the angle DAC in the alternate segment of the circle. [III. 32]

4.prop.10.p1181

Since, then, the angle BDC is equal to the angle DAC, let the angle CDA be added to each; therefore the whole angle BDA is equal to the two angles CDA, DAC.

4.prop.10.p1182

But the exterior angle BCD is equal to the angles CDA, DAC; [I. 32] therefore the angle BDA is also equal to the angle BCD.

4.prop.10.p1183

But the angle BDA is equal to the angle CBD, since the side AD is also equal to AB; [I. 5] so that the angle DBA is also equal to the angle BCD.

4.prop.10.p1184

Therefore the three angles BDA, DBA, BCD are equal to one another.

4.prop.10.p1185

And, since the angle DBC is equal to the angle BCD, the side BD is also equal to the side DC. [I. 6]

4.prop.10.p1186

But BD is by hypothesis equal to CA; therefore CA is also equal to CD, so that the angle CDA is also equal to the angle DAC; [I. 5] therefore the angles CDA, DAC are double of the angle DAC.

4.prop.10.p1187

But the angle BCD is equal to the angles CDA, DAC; therefore the angle BCD is also double of the angle CAD.

4.prop.10.p1188

But the angle BCD is equal to each of the angles BDA, DBA; therefore each of the angles BDA, DBA is also double of the angle DAB.

4.prop.10.p1189

Therefore the isosceles triangle ABD has been constructed having each of the angles at the base DB double of the remaining one. Q. E. F.

4.prop.11.p1190

In a given circle to inscribe an equilateral and equiangular pentagon.

4.prop.11.p1191

Let ABCDE be the given circle; thus it is required to inscribe in the circle ABCDE an equilateral and equiangular pentagon.

4.prop.11.p1192

Let the isosceles triangle FGH be set out having each of the angles at G, H double of the angle at F; [IV. 10] let there be inscribed in the circle ABCDE the triangle ACD equiangular with the triangle FGH, so that the angle CAD is equal to the angle at F and the angles at G, H respectively equal to the angles ACD, CDA; [IV. 2] therefore each of the angles ACD, CDA is also double of the angle CAD.

4.prop.11.p1193

Now let the angles ACD, CDA be bisected respectively by the straight lines CE, DB [I. 9], and let AB, BC, DE, EA be joined.

4.prop.11.p1194

Then, since each of the angles ACD, CDA is double of the angle CAD, and they have been bisected by the straight lines CE, DB, therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another.

4.prop.11.p1195

But equal angles stand on equal circumferences; [III. 26] therefore the five circumferences AB, BC, CD, DE, EA are equal to one another.

4.prop.11.p1196

But equal circumferences are subtended by equal straight lines; [III. 29] therefore the five straight lines AB, BC, CD, DE, EA are equal to one another; therefore the pentagon ABCDE is equilateral.

4.prop.11.p1197

I say next that it is also equiangular.

4.prop.11.p1198

For, since the circumference AB is equal to the circumference DE, let BCD be added to each; therefore the whole circumference ABCD is equal to the whole circumference EDCB.

4.prop.11.p1199

And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the angle BAE is also equal to the angle AED. [III. 27]

4.prop.11.p1200

For the same reason each of the angles ABC, BCD, CDE is also equal to each of the angles BAE, AED; therefore the pentagon ABCDE is equiangular.

4.prop.11.p1201

But it was also proved equilateral; therefore in the given circle an equilateral and equiangular pentagon has been inscribed. Q. E. F.

4.prop.12.p1202

About a given circle to circumscribe an equilateral and equiangular pentagon.

4.prop.12.p1203

Let ABCDE be the given circle; thus it is required to circumscribe an equilateral and equiangular pentagon about the circle ABCDE.

4.prop.12.p1204

Let A, B, C, D, E be conceived to be the angular points of the inscribed pentagon, so that the circumferences AB, BC, CD, DE, EA are equal; [IV. 11] through A, B, C, D, E let GH, HK, KL, LM, MG be drawn touching the circle; [III. 16, Por.] let the centre F of the circle ABCDE be taken [III. 1], and let FB, FK, FC, FL, FD be joined.

4.prop.12.p1205

Then, since the straight line KL touches the circle ABCDE at C, and FC has been joined from the centre F to the point of contact at C, therefore FC is perpendicular to KL; [III. 18] therefore each of the angles at C is right.

4.prop.12.p1206

For the same reason the angles at the points B, D are also right.

4.prop.12.p1207

And, since the angle FCK is right, therefore the square on FK is equal to the squares on FC, CK.

4.prop.12.p1208

For the same reason [I. 47] the square on FK is also equal to the squares on FB, BK; so that the squares on FC, CK are equal to the squares on FB, BK, of which the square on FC is equal to the square on FB; therefore the square on CK which remains is equal to the square on BK.

4.prop.12.p1209

Therefore BK is equal to CK.

4.prop.12.p1210

And, since FB is equal to FC, and FK common, the two sides BF, FK are equal to the two sides CF, FK; and the base BK equal to the base CK; therefore the angle BFK is equal to the angle KFC, [I. 8] and the angle BKF to the angle FKC. Therefore the angle BFC is double of the angle KFC, and the angle BKC of the angle FKC.

4.prop.12.p1211

For the same reason the angle CFD is also double of the angle CFL, and the angle DLC of the angle FLC.

4.prop.12.p1212

Now, since the circumference BC is equal to CD, the angle BFC is also equal to the angle CFD. [III. 27]

4.prop.12.p1213

And the angle BFC is double of the angle KFC, and the angle DFC of the angle LFC; therefore the angle KFC is also equal to the angle LFC.

4.prop.12.p1214

But the angle FCK is also equal to the angle FCL; therefore FKC, FLC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them; therefore they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle; [I. 26] therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC.

4.prop.12.p1215

And, since KC is equal to CL, therefore KL is double of KC.

4.prop.12.p1216

For the same reason it can be proved that HK is also double of BK.

4.prop.12.p1217

And BK is equal to KC; therefore HK is also equal to KL.

4.prop.12.p1218

Similarly each of the straight lines HG, GM, ML can also be proved equal to each of the straight lines HK, KL; therefore the pentagon GHKLM is equilateral.

4.prop.12.p1219

I say next that it is also equiangular.

4.prop.12.p1220

For, since the angle FKC is equal to the angle FLC, and the angle HKL was proved double of the angle FKC, and the angle KLM double of the angle FLC, therefore the angle HKL is also equal to the angle KLM.

4.prop.12.p1221

Similarly each of the angles KHG, HGM, GML can also be proved equal to each of the angles HKL, KLM; therefore the five angles GHK, HKL, KLM, LMG, MGH are equal to one another.

4.prop.12.p1222

Therefore the pentagon GHKLM is equiangular.

4.prop.12.p1223

And it was also proved equilateral; and it has been circumscribed about the circle ABCDE. Q. E. F.

4.prop.13.p1224

In a given pentagon, which is equilateral and equiangular, to inscribe a circle.

4.prop.13.p1225

Let ABCDE be the given equilateral and equiangular pentagon; thus it is required to inscribe a circle in the pentagon ABCDE.

4.prop.13.p1226

For let the angles BCD, CDE be bisected by the straight lines CF, DF respectively; and from the point F, at which the straight lines CF, DF meet one another, let the straight lines FB, FA, FE be joined.

4.prop.13.p1227

Then, since BC is equal to CD, and CF common, the two sides BC, CF are equal to the two sides DC, CF; and the angle BCF is equal to the angle DCF; therefore the base BF is equal to the base DF, and the triangle BCF is equal to the triangle DCF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4]

4.prop.13.p1228

Therefore the angle CBF is equal to the angle CDF.

4.prop.13.p1229

And, since the angle CDE is double of the angle CDF, and the angle CDE is equal to the angle ABC, while the angle CDF is equal to the angle CBF; therefore the angle CBA is also double of the angle CBF; therefore the angle ABF is equal to the angle FBC; therefore the angle ABC has been bisected by the straight line BF.

4.prop.13.p1230

Similarly it can be proved that the angles BAE, AED have also been bisected by the straight lines FA, FE respectively.

4.prop.13.p1231

Now let FG, FH, FK, FL, FM be drawn from the point F perpendicular to the straight lines AB, BC, CD, DE, EA.

4.prop.13.p1232

Then, since the angle HCF is equal to the angle KCF, and the right angle FHC is also equal to the angle FKC, FHC, FKC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore the perpendicular FH is equal to the perpendicular FK.

4.prop.13.p1233

Similarly it can be proved that each of the straight lines FL, FM, FG is also equal to each of the straight lines FH, FK; therefore the five straight lines FG, FH, FK, FL, FM are equal to one another.

4.prop.13.p1234

Therefore the circle described with centre F and distance one of the straight lines FG, FH, FK, FL, FM will pass also through the remaining points; and it will touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right.

4.prop.13.p1235

For, if it does not touch them. but cuts them, it will result that the straight line drawn at right angles to the diameter of the circle from its extremity falls within the circle: which was proved absurd. [III. 16]

4.prop.13.p1236

Therefore the circle described with centre F and distance one of the straight lines FG, FH, FK, FL, FM will not cut the straight lines AB, BC, CD, DE, EA; therefore it will touch them.

4.prop.13.p1237

Let it be described, as GHKLM.

4.prop.13.p1238

Therefore in the given pentagon, which is equilateral and equiangular, a circle has been inscribed. Q. E. F.

4.prop.14.p1239

About a given pentagon, which is equilateral and equiangular, to circumscribe a circle.

4.prop.14.p1240

Let ABCDE be the given pentagon, which is equilateral and equiangular; thus it is required to circumscribe a circle about the pentagon ABCDE.

4.prop.14.p1241

Let the angles BCD, CDE be bisected by the straight lines CF, DF respectively, and from the point F, at which the straight lines meet, let the straight lines FB, FA, FE be joined to the points B, A, E.

4.prop.14.p1242

Then in manner similar to the preceding it can be proved that the angles CBA, BAE, AED have also been bisected by the straight lines FB, FA, FE respectively.

4.prop.14.p1243

Now, since the angle BCD is equal to the angle CDE, and the angle FCD is half of the angle BCD, and the angle CDF half of the angle CDE, therefore the angle FCD is also equal to the angle CDF, so that the side FC is also equal to the side FD. [I. 6]

4.prop.14.p1244

Similarly it can be proved that each of the straight lines FB, FA, FE is also equal to each of the straight lines FC, FD; therefore the five straight lines FA, FB, FC, FD, FE are equal to one another.

4.prop.14.p1245

Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC, FD, FE will pass also through the remaining points, and will have been circumscribed.

4.prop.14.p1246

Let it be circumscribed, and let it be ABCDE.

4.prop.14.p1247

Therefore about the given pentagon, which is equilateral and equiangular, a circle has been circumscribed. Q. E. F.

4.prop.15.p1248

In a given circle to inscribe an equilateral and equiangular hexagon.

4.prop.15.p1249

Let ABCDEF be the given circle; thus it is required to inscribe an equilateral and equiangular hexagon in the circle ABCDEF.

4.prop.15.p1250

Let the diameter AD of the circle ABCDEF be drawn; let the centre G of the circle be taken, and with centre D and distance DG let the circle EGCH be described; let EG, CG be joined and carried through to the points B, F, and let AB, BC, CD, DE, EF, FA be joined.

4.prop.15.p1251

I say that the hexagon ABCDEF is equilateral and equiangular.

4.prop.15.p1252

For, since the point G is the centre of the circle ABCDEF, GE is equal to GD.

4.prop.15.p1253

Again, since the point D is the centre of the circle GCH, DE is equal to DG.

4.prop.15.p1254

But GE was proved equal to GD; therefore GE is also equal to ED; therefore the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one another, inasmuch as, in isosceles triangles, the angles at the base are equal to one another. [I. 5]

4.prop.15.p1255

And the three angles of the triangle are equal to two right angles; [I. 32] therefore the angle EGD is one-third of two right angles.

4.prop.15.p1256

Similarly, the angle DGC can also be proved to be onethird of two right angles.

4.prop.15.p1257

And, since the straight line CG standing on EB makes the adjacent angles EGC, CGB equal to two right angles, therefore the remaining angle CGB is also one-third of two right angles.

4.prop.15.p1258

Therefore the angles EGD, DGC, CGB are equal to one another; so that the angles vertical to them, the angles BGA, AGF, FGE are equal. [I. 15]

4.prop.15.p1259

Therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another.

4.prop.15.p1260

But equal angles stand on equal circumferences; [III. 26] therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another.

4.prop.15.p1261

And equal circumferences are subtended by equal straight lines; [III. 29] therefore the six straight lines are equal to one another; therefore the hexagon ABCDEF is equilateral.

4.prop.15.p1262

I say next that it is also equiangular.

4.prop.15.p1263

For, since the circumference FA is equal to the circumference ED, let the circumference ABCD be added to each; therefore the whole FABCD is equal to the whole EDCBA; and the angle FED stands on the circumference FABCD, and the angle AFE on the circumference EDCBA; therefore the angle AFE is equal to the angle DEF. [III. 27]

4.prop.15.p1264

Similarly it can be proved that the remaining angles of the hexagon ABCDEF are also severally equal to each of the angles AFE, FED; therefore the hexagon ABCDEF is equiangular.

4.prop.15.p1265

But it was also proved equilateral; and it has been inscribed in the circle ABCDEF.

4.prop.15.p1266

Therefore in the given circle an equilateral and equiangular hexagon has been inscribed. Q. E. F.

4.prop.15.p1267

Porism. From this it is manifest that the side of the hexagon is equal to the radius of the circle.

4.prop.15.p1268

And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle an equilateral and equiangular hexagon in conformity with what was explained in the case of the pentagon.

4.prop.15.p1269

And further by means similar to those explained in the case of the pentagon we can both inscribe a circle in a given hexagon and circumscribe one about it. Q. E. F.

4.prop.16.p1270

In a given circle to inscribe a fifteen-angled figure which shall be both equilateral and equiangular.

4.prop.16.p1271

Let ABCD be the given circle; thus it is required to inscribe in the circle ABCD a fifteenangled figure which shall be both equilateral and equiangular.

4.prop.16.p1272

In the circle ABCD let there be inscribed a side AC of the equilateral triangle inscribed in it, and a side AB of an equilateral pentagon; therefore, of the equal segments of which there are fifteen in the circle ABCD, there will be five in the circumference ABC which is one-third of the circle, and there will be three in the circumference AB which is one-fifth of the circle; therefore in the remainder BC there will be two of the equal segments.

4.prop.16.p1273

Let BC be bisected at E; [III. 30] therefore each of the circumferences BE, EC is a fifteenth of the circle ABCD.

4.prop.16.p1274

If therefore we join BE, EC and fit into the circle ABCD straight lines equal to them and in contiguity, a fifteen-angled figure which is both equilateral and equiangular will have been inscribed in it. Q. E. F.

4.prop.16.p1275

And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle a fifteen-angled figure which is equilateral and equiangular.

4.prop.16.p1276

And further, by proofs similar to those in the case of the pentagon, we can both inscribe a circle in the given fifteenangled figure and circumscribe one about it. Q. E. F.

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