For, since the angle FKC is equal to the angle FLC,
and that the angle HKL is double of the angle FKC, and the angle KLM double of the angle FLC, as was shewn,
therefore the angle HKL is equal to the angle KLM. [Axiom 6.
In the same manner it may be shewn that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM;
therefore the pentagon GHKLM is equiangular. And it has been shewn to be equilateral.
Wherefore an equilateral and equiangular pentagon has been described about the given circle. q.e.f.
PROPOSITION 13. PROBLEM.
To inscribe a circle in a given equilateral and equiangular pentagon.
Let ABCDE be the given equilateral and equiangular pentagon: it is required to inscribe a circle in the pentagon ABCDE.
Bisect the angles BCD, CDE by the straight lines CF,DF; [1.9. and from the point F, at which they meet, draw the straight lines FB, FA, FE.
Then, because BC is equal to DC [Hypothesis.
and CF is common to the two triangles BCF, DCF;
the two sides BC, CF are equal to the two sides DC, CF, each to each;
and the angle BCF is equal to the angle DCF; [Constr.
therefore the base BF is equal to the base DF, and the
