To inscribe a triangle equiangular with a given triangle in a given circle.

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.

Q.E.F.