Details

The situation is illustrated in step 1, where the line segment is a diameter of the incircle. The point is determined by extending the segment . The lemma states that . Drag the Locator at to modify the shape of the triangle.

The proof proceeds by making a translation of the incircle along line , so that point is now placed on , as shown in step 2. A dilation is now applied to the translated circle, keeping its top point fixed, until the dilated circle is tangent to the extended sides and (move the slider until the tangent points and appear in the figure). This is one of the excircles of the triangle.

This excircle and the incircle determine three vertices where the tangents property holds (i.e., two tangents to a circle from an external point are equal), namely (tangents to the excircle and , and tangents to the incircle and ); (tangents to the excircle and , and tangents to the incircle and ); and (tangents to the excircle and , and tangents to the incircle and ).

From here on it is a matter of simple substitutions:

Starting from :

,

,

,

,

,

,

, Q.E.D.

Reference

[1] Y. Zhao. "Three Lemmas in Geometry." Massachusetts Institute of Technology Winter Camp 2010. yufeizhao.com/olympiad/three_geometry_lemmas.pdf.