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 2 of 4: Show the incircle touchpoints P and Q on AC are symmetric and match the excircle touchpoints

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In plain words

The standard tangent-length formula for $\omega_1$ on $\triangle ABC$ and $\omega_2$ on $\triangle ADC$ , combined with $AB - BC = CD - AD$ from Step 1, gives $AP = CQ$ .

AP = \frac{AC+AB-BC}{2} = \frac{CA+CD-AD}{2} = CQ

Detailed analysis

Let $\omega_1$ and $\omega_2$ touch $AC$ at $P$ and $Q$ . Standard tangent lengths in $\triangle ABC$ and $\triangle ADC$ give $AP = \dfrac{AC+AB-BC}{2}$ and $CQ = \dfrac{CA+CD-AD}{2}$ ; since $AB - BC = CD - AD$ by Step 1, we obtain $AP = CQ$ (with $P \ne Q$ as $BA \ne BC$ ). By the standard property of excircles, $Q$ is also the tangency point on $AC$ of the excircle of $\triangle ABC$ opposite $B$ , and $P$ is the tangency point on $AC$ of the excircle of $\triangle ADC$ opposite $D$ .