Construction of Excircle to Triangle

Theorem

It is possible to construct an excircle to a triangle tangent to any of its three sides.

Let $ABC$ be the given triangle.

Let us be required to construct the excircle tangent to $BC$.

Let $AB$ and $AC$ be produced beyond $B$ and $C$ to $X$ and $Y$ respectively.

Let $\angle CBX$ and $\angle BCY$ be bisected by $BE$ and $CE$ and let these lines join at $E$.

From $E$ construct the perpendiculars $EP, EQ, ER$ to $AY, AX, BC$ respectively.

Construct the circle with radius $ER$ and center $E$.

This is the required excircle.

We have that $\angle XBE = \angle CBE$ and $\angle BQE = \angle BRE$, a right angle, and $BE$ is common.

So from Triangle Angle-Side-Angle Congruence we have that $\triangle EQB = \triangle ERB$.

So $EQ = ER$.

For the same reason $ER = EP$.

So $EQ = ER = EP$.

So the circle drawn with radius $ER$ will pass through $P, Q$ and $R$.

From Line at Right Angles to Diameter of Circle it follows that $AX, AY, BC$ are tangent to the circle $PQR$.

Hence the result.

$\blacksquare$