be equal to the angle at F, and each of the angles ACD, equal to the angle at G or H [IV. 2.
and therefore each of angles ACD,ADC is double of the angle CAD; bisect the angles ACD, ADC by the straight line CE,DB; [1.9.
and join AB, BC, AE, ED.
ABCDE shall be the pentagon required.
For because each of the angles ACD, ADC is double of the angle CAD,
and that they are bisected by the straight lines CE, DB,
therefore the five angles ADB, BDC, CAD, DCE, ECA
equal to one another.
equal angles stand on equal arcs; [III. 26.
therefore the five arcs AB, BC, CD, DE, EA are equal to one another.
And equal arcs are subtended by equal straight lines; [III. 29.
therefore the five straight lines AB, BC, CD, DE, EA are equal to one another;
and therefore the pentagon ABCDE is equilateral.
It is also equiangular.
For, the arc AB is equal to the arc DE;
each of these add the arc BCD;
therefore the whole arc ABCD is equal to the whole arc BCDE. [Axiom 2.
And the angle AED stands on the arc ABCD, and the angle BAE on the arc BCDE. Therefore the angle AED is equal to the angle BAE. [III. 27.
For the same reason each of the angles ABC, BCD, DE is equal to the angle AED or BAE;
Therefore the pentagon ABCDE is equiangular.
And it has been shewn to be equilateral.
Wherefore an equilateral and equiangular pentagon has inscribed in the given circle, q.e.f.
