Inscribing in Circle Triangle Equiangular with Given

Construction
Let $ABC$ be the given circle and let $\triangle DEF$ be the given triangle.


 * Euclid-IV-2.png

Let $GH$ be drawn tangent to $ABC$ at $A$.

On the straight line $AH$ construct $\angle HAC$ equal to $\angle DEF$.

On the straight line $AG$ construct $\angle GAB$ equal to $\angle DFE$.

Join $BC$.

Then $\triangle ABC$ is the given triangle.

Proof
We have that $AH$ touches the circle $ABC$, and $AC$ is a chord of circle $ABC$.

Then from Angles made by Chord with Tangent‎ $\angle HAC = \angle ABC$.

But $\angle HAC = \angle DEF$.

Similarly, we have that $AG$ touches the circle $ABC$, and $AB$ is a chord of circle $ABC$.

Then from Angles made by Chord with Tangent‎ $\angle GAB = \angle ACB$.

But $\angle GAB = \angle DFE$.

So from Sum of Angles of Triangle Equals Two Right Angles it follows that the remaining angles are equal also: $\angle BAC = \angle EDF$.

Hence the result.