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 4 of 4: Identify the common intersection of P'Q and PQ' with the point K on omega

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In plain words

Because $\omega$ is inscribed in $\angle ABC$ , the homothety at $B$ taking the $B$ -excircle to $\omega$ carries $Q$ to the point $K \in \omega$ where the tangent is parallel to $AC$ , and the homothety at $D$ taking the $D$ -excircle of $\triangle ADC$ to $\omega$ carries $P$ to the same point $K$ ; then $K = P'Q \cap PQ'$ sends diameter $P'P$ to the parallel diameter $QQ'$ , making $K$ the exsimilicenter of $\omega_1$ and $\omega_2$ .

K \in \omega \text{ with tangent } \parallel AC \implies B, P', Q, K \text{ collinear and } D, Q', P, K \text{ collinear} \implies K = P'Q \cap PQ' \text{ is the exsimilicenter of } \omega_1, \omega_2

Detailed analysis

Let $K$ be the point of $\omega$ closest to $AC$ at which the tangent to $\omega$ is parallel to $AC$ . Since $\omega$ and the $B$ -excircle of $\triangle ABC$ are both inscribed in $\angle ABC$ , the homothety at $B$ between them maps $Q$ to $K$ , so $K$ lies on line $BP'Q$ ; similarly, since $\omega$ and the $D$ -excircle of $\triangle ADC$ are both tangent to lines $AD$ and $CD$ , the homothety at $D$ between them maps $P$ to $K$ , so $K$ also lies on line $DQ'P$ . Because $P'P \parallel QQ'$ are parallel diameters of $\omega_1$ and $\omega_2$ with $K = P'Q \cap PQ'$ matching same-side endpoints ( $P'$ with $Q$ and $P$ with $Q'$ ), the homothety centered at $K$ maps diameter $P'P$ to diameter $QQ'$ with positive ratio, hence maps $\omega_1$ to $\omega_2$ . Therefore $K \in \omega$ is the external center of similitude of $\omega_1$ and $\omega_2$ , where their common external tangents intersect.