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 1 of 4: Prove the Pitot-type relation AB + AD = CB + CD from omega

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In plain words

Expressing each side of $ABCD$ in terms of the tangent segments from the four vertices to $\omega$ cancels in pairs.

BK=BL,\quad AK=AN,\quad CL=CM,\quad DN=DM \implies AB+AD=CB+CD

Detailed analysis

Let $\omega$ touch ray $BA$ beyond $A$ at $K$ , ray $BC$ beyond $C$ at $L$ , and the extensions of $AD$ and $CD$ beyond $D$ at $N$ and $M$ . Equal tangents from each vertex to $\omega$ give $BK = BL$ , $AK = AN$ , $CL = CM$ , and $DN = DM$ . Therefore $AB + AD = (BK - AK) + (AN - DN) = BK - DN = BL - DM = (BL - CL) + (CM - DM) = CB + CD$ .