Therefore the angle BDA is equal to the angle BCD [[Ax .l. But the angle BDA is equal to the angle DBA, [I. 5.
because AD is equal to AB.
Therefore each of the angles BDA, DBA, is equal to the angle BCD. [Axiom 6.
And, because the angle DBC is equal to the angle BCD, the side DB is equal to the side DC; [1.6.
but DB was made equal to CA;
therefore CA is equal to CD, [Axiom 6.
and therefore the angle CAD is equal to the angle CDA. [I.5
Therefore the angles CAD, CDA are together double the angle CAD.
But the angle BCD is equal to the angles CAD, CDA. [I.32
Therefore the angle BCD is double of the angle CAD,
And the angle BCD has been shewn to be equal to each of the angles BDA, DBA;
therefore each of the angles BDA, DBA is double of the angle BAD.
Wherefore an isosceles triangle has been described, having each of the angles at the base double of the third angle, q.e.f.
PROPOSITION 11. PROBLEM.
To inscribe an equilateral and equiangular pentagon in a given circle.
Let ABCDE be the given circle: it is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE.
Describe an isosceles triangle, FGH, having each of the angles at G, H, double of the angle at F [IV. 10.
in the circle ABCDE, inscribe the triangle ACD, equiangular to the triangle FGH, so that the angle CAD may
