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 6 of 6: Combine both rewrites and add

TX=YZ

Previous step Next step

RT+XP=ZQ+SY

Detailed analysis

The congruences from the second step give $ZQ=RT$ and $XP=SY$ , so $RT+XP=ZQ+SY$ , which shows $AD+DT+XA=CD+DY+ZC$ . Adding the equal quantities $TX=YZ$ from the third step to both sides finally gives $AD+DT+TX+XA=CD+DY+YZ+ZC$ , as required.