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For the same reason it can be proved that

HK is also double of BK.

And BK is equal to KC;

therefore HK is also equal to KL.

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.

I say next that it is also equiangular.

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.

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.

Therefore the pentagon GHKLM is equiangular.

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

PROPOSITION 13.

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

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

For let the angles BCD, CDE be bisected by the straight lines CF, DF respectively; and from the point F, at

Euclid. Euclid's Elements. Sir Thomas Little Heath. New York. Dover. 1956.

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