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. [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. [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.
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.
Similarly we can prove that AC has also been cut in extreme and mean ratio at H, and its greater segment CH is equal to the side of the pentagon.
Q. E. D.
Complete text
Book I