Let ABC be the given circle, and DEF the given triangle.
It is required to inscribe a triangle equiangular with the triangle DEF in the circle ABC.
Draw GH touching the circle ABC at A.
Construct the angle HAC equal to the angle DEF on the straight line AH and at the point A on it, and construct the angle GAB equal to the angle DFE on the straight line AG and at the point A on it. Join BC.
Then, since a straight line AH touches the circle ABC, and from the point of contact at A the straight line AC is drawn across in the circle, therefore the angle HAC equals the angle ABC in the alternate segment of the circle.
But the angle HAC equals the angle DEF, therefore the angle ABC also equals the angle DEF.
For the same reason the angle ACB also equals the angle DFE, therefore the remaining angle BAC also equals the remaining angle EDF.
Therefore a triangle equiangular with the given triangle has been inscribed in the given circle.