# Construction of Thébault Circles

### What Might This Be About?

### Problem

Given $\Delta ABC$ and point $D$ on $BC.$

Construct a circle tangent to $AD,$ $CD,$ and the circumcircle of $\Delta ABC.$

### Solution

Solution proceeds in several steps:

Find the incenter $I$ of $\Delta ABC.$

Drop a perpendicular from $I$ to the bisector of $\angle ADC.$ Mark its intersection with $BC$ - $F.$

- p>Erect a perpendicular to $BC$ at $F$. Mark $K$ - its intersection with the bisector of $\angle ADC.$

$K$ is the center of Thébault's circle.

For a proof, recollect Y. Sawayama's Lemma: If $F$ is the point of tangency of the Thébault circle $(K)$ and $BC$ then $IF$ passes through its point of tangency with $AD,$ say $H.$ But $FH$ is perpendicular to the bisector of $\angle ADC$ into which $(K)$ is inscribed.

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny66206492