If in an equilateral and equiangular pentagon straight lines subtend two angles taken in order, they cut one another in extreme and mean ratio, and their greater segments are equal to the side of the pentagon.
在等边等角五边形中，依次连接两个角所引的对角线互相分割成中外比，且较长的线段等于五边形的边。
正五边形 ABCDE 内接于圆，圆心 O；对角线 AC 与 BE 交于 H。则 AH=BH 等于五边形的边长，且对角线被 H 分割成中外比（黄金分割）。
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For in the equilateral and equiangular pentagon ABCDE let the straight lines AC, BE, cutting one another at the point H, subtend two angles taken in order, the angles at A, B; I say that each of them has been cut in extreme and mean ratio at the point H, and their greater segments are equal to the side of the pentagon. For let the circle ABCDE be circumscribed about the pentagon ABCDE. [IV. 14] Then, since the two straight lines EA, AB are equal to the two AB, BC, and they contain equal angles, therefore the base BE is equal to the base AC, the triangle ABE is equal to the triangle ABC, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend.
作五边形ABCDE的外接圆。由于EA、AB分别等于AB、BC且夹角相等，故底边BE等于AC，三角形ABE全等于三角形ABC，对应角相等。
[I. 4] Therefore the angle BAC is equal to the angle ABE; therefore the angle AHE is double of the angle BAH. [I. 32] But the angle EAC is also double of the angle BAC, inasmuch as the circumference EDC is also double of the circumference CB; [III. 28, VI. 33] therefore the angle HAE is equal to the angle AHE; hence the straight line HE is also equal to EA, that is, to AB. [I. 6] And, since the straight line BA is equal to AE, the angle ABE is also equal to the angle AEB.
由角BAC等于角ABE，得角AHE等于角BAH的两倍。又角EAC等于角BAC的两倍，故角HAE等于角AHE，从而HE等于EA，即等于AB。
[I. 5] But the angle ABE was proved equal to the angle BAH; therefore the angle BEA is also equal to the angle BAH. And the angle ABE is common to the two triangles ABE and ABH; therefore the remaining angle BAE is equal to the remaining angle AHB; [I. 32] therefore the triangle ABE is equiangular with the triangle ABH; therefore, proportionally, as EB is to BA, so is AB to BH. [VI. 4] But BA is equal to EH; therefore, as BE is to EH, so is EH to HB.
由BA等于AE，得角ABE等于角AEB。已证角ABE等于角BAH，故角BEA等于角BAH。公共角ABE，故三角形ABE与ABH相似，对应边成比例：EB比BA等于AB比BH。
And BE is greater than EH; therefore EH is also greater than HB. [V. 14] Therefore BE has been cut in extreme and mean ratio at H, and the greater segment HE is equal to the side of the pentagon.
因BA等于EH，故BE比EH等于EH比HB。由BE大于EH，得EH大于HB。因此BE在H点被分成中外比，较长线段HE等于五边形的边。