# Inscribing in Circle Triangle Equiangular with Given

## Theorem

In the words of Euclid:

*In a given circle to inscribe a triangle equiangular with a given triangle.*

(*The Elements*: Book $\text{IV}$: Proposition $2$)

## Construction

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

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.

$\blacksquare$

## Historical Note

This theorem is Proposition $2$ of Book $\text{IV}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 2*(2nd ed.) ... (previous) ... (next): Book $\text{IV}$. Propositions - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)