In a given circle to inscribe a triangle equiangular with a given triangle.
Proof
Let ABC be the given circle and DEF the given triangle. Draw the
tangent GH at any point A of the circle (III.16, III.17). On GH
construct ∠HAC=∠DEF (I.23), and ∠GAB=∠DFE. Join BC. By III.32 (tangent–chord angle equals the
inscribed angle in the alternate segment), ∠ACB=∠HAB=∠DEF and ∠ABC=∠GAC=∠DFE. By I.32 the
remaining angles agree; so △ABC is equiangular with
△DEF.