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.