Problem 4

Let

\Gamma

be a circle with centre

I

, and

ABCD

a convex quadrilateral such that each of the segments

AB

BC

CD

and

DA

is tangent to

\Gamma

. Let

\Omega

be the circumcircle of the triangle

AIC

. The extension of

BA

beyond

A

meets

\Omega

X

, and the extension of

BC

beyond

C

meets

\Omega

Z

. The extensions of

AD

and

CD

beyond

D

meet

\Omega

Y

and

T

, respectively. Prove that

AD+DT+TX+XA=CD+DY+YZ+ZC.

Step 2 of 6: Find the reflection symmetry

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In plain words

The incenter axis of the two circles exchanges the opposite points on $\Omega$ .

X\leftrightarrow Y,\quad T\leftrightarrow Z\quad\text{under reflection in }OI

Detailed analysis

Let $O$ be the centre of $\Omega$ . Since $I$ is the incenter of the tangential quadrilateral, an angle chase in cyclic quadrilateral $ACZX$ gives $\angle IZX=\angle IAB=\angle IAD=\angle IAY$ . Thus the chords $IX$ and $IY$ are equal, so $X,Y$ are mirror images in $OI$ . The same argument on $C,I,T,Z$ gives that $T,Z$ are mirror images in $OI$ .