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 5 of 6: Rewrite

CD+DY+ZC

the same way

Previous step Next step

CD+DY+ZC=SY+ZQ

Detailed analysis

Symmetrically, $CD=CR+RD=CQ+DS$ , and the orderings $A,S,D,Y$ on line $AD$ and $B,Q,C,Z$ on line $BC$ give $DY=SY-SD$ and $ZC=ZQ-QC$ . Adding these gives $CD+DY+ZC=(CD-DS-QC)+SY+ZQ=SY+ZQ$ , because $CD-DS-QC=0$ .