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 4 of 6: Rewrite

AD+DT+XA

using tangent lengths

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AD+DT+XA=RT+XP

Detailed analysis

Since $S$ lies between $A$ and $D$ on side $DA$ , $AD=AS+SD=AP+DR$ ; since $C,R,D,T$ lie in this order on line $CD$ , $DT=RT-RD$ ; and since $X,A,P,B$ lie in this order on line $AB$ , $XA=XP-AP$ . Adding these gives $AD+DT+XA=(AD-DR-AP)+RT+XP=RT+XP$ , because $AD-DR-AP=0$ .