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 3 of 6: Deduce

TX=YZ

Previous step Next step

TX=YZ

Detailed analysis

The reflection in $OI$ preserves distances and swaps $T\leftrightarrow Z$ , $X\leftrightarrow Y$ , so it sends segment $TX$ to segment $ZY$ . Therefore $TX=YZ$ .