Problem 6

Let

ABCD

be a convex quadrilateral with

BA \ne BC

. Denote the incircles of triangles

ABC

and

ADC

\omega_1

and

\omega_2

respectively. Suppose that there exists a circle

\omega

tangent to ray

BA

beyond

A

and to the ray

BC

beyond

C

, which is also tangent to the lines

AD

and

CD

. Prove that the common external tangents to

\omega_1

and

\omega_2

intersect on

\omega

Step 3 of 4: Use homothety at B and at D to align the antipodal diameter endpoints P' and Q'

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In plain words

The homothety at $B$ mapping $\omega_1$ to the $B$ -excircle of $\triangle ABC$ carries the top of $\omega_1$ (where the tangent is parallel to $AC$ ) to the touchpoint $Q$ of the $B$ -excircle on $AC$ .

PP' \perp AC \text{ diameter of } \omega_1,\quad QQ' \perp AC \text{ diameter of } \omega_2 \implies B, P', Q \text{ collinear and } D, Q', P \text{ collinear}

Detailed analysis

Let $PP'$ and $QQ'$ be the diameters of $\omega_1$ and $\omega_2$ perpendicular to $AC$ . In $\triangle ABC$ , the tangent to $\omega_1$ at $P'$ is parallel to $AC$ , so the positive homothety centered at $B$ taking $\omega_1$ to the $B$ -excircle of $\triangle ABC$ carries $P'$ to the point where that excircle touches $AC$ , which by Step 2 is $Q$ ; hence $B, P', Q$ are collinear. By the exact same argument in $\triangle ADC$ , $D, Q', P$ are collinear.