Circumscribing about Circle a Triangle Equiangular with Given

Theorem

About a given circle, it is possible to circumscribe a triangle which is equiangular to a given triangle.

Construction

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

Produce $EF$ in both directions to $G$ and $H$.

Find the center $K$ of $ABC$, and draw $KB$ at random.

Construct $\angle BKA$ equal to $\angle DEG$.

Construct $\angle BKC$ equal to $\angle DFH$.

Through $A, B, C$ draw tangents $LM, MN, LN$ to $ABC$.

Then $\triangle LMN$ is the required triangle.

Proof

From Radius at Right Angle to Tangent, $LM$ is perpendicular to $AK$, $MN$ is perpendicular to $BK$, and $LN$ is perpendicular to $CK$.

We have that $\angle KAM, \angle KBM$ are right angles.

This leaves us with $\angle AMB + \angle AKB$ equal to two right angles.

We have that $\angle AKB = \angle DEG$ and so $\angle AMB + \angle DEG$ equals two right angles.

But from Two Angles on a Straight Line make Two Right Angles $\angle DEF + \angle DEG$ equals two right angles.

It follows that $\angle AMB = \angle DEF$.

In a similar way we can show that $\angle ALC = \angle EDF$ and $\angle BNC = \angle DFE$.

Hence the result.

$\blacksquare$

Historical Note

This is Proposition 3 of Book IV of Euclid's The Elements.