Problem. In a given circle (ABC) to inscribe a triangle equiangular to a given triangle (DEF).
Sol.—Take any point A in the circumference, and at it draw the tangent GH; then make the angle HAC equal to E, and GAB equal to F [I. xxiii.] Join BC. ABC is a triangle fulfilling the required conditions.
Dem.—The angle E is equal to HAC (const.), and HAC is equal to the angle ABC in the alternate segment [III. xxxii.]. Hence the angle E is equal to ABC. In like manner the angle F is equal to ACB. Therefore [I. xxxii.] the remaining angle D is equal to BAC. Hence the triangle ABC inscribed in the given circle is equiangular to DEF.