Pedal Circle of Excenter is Excircle

Theorem

Let $\triangle ABC$ be a triangle.

Let $H$ be an excenter of $\triangle ABC$.

Then the pedal circle of $H$ is the excircle of $\triangle ABC$ whose center is $H$.

Proof

Let $\EE$ denote the excircle of $\triangle ABC$ whose center is $H$.

Aiming for a contradiction, suppose $\EE$ is not the pedal circle of $H$.

By definition, the sides of $\triangle ABC$ are tangent to $\EE$.

Hence from Line at Right Angles to Diameter of Circle: Porism, the line through $H$ to those point of tangency to $\EE$ are perpendicular to the sides.

Hence $\EE$ is the pedal circle of $H$ by definition.

Hence by Proof by Contradiction, the pedal circle of $H$ is the excircle of $\triangle ABC$ whose center is $H$.

$\blacksquare$

Also see

• Pedal Circle of Incenter is Incircle