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 1 of 6: Name the incircle's tangent points

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In plain words

The two tangent segments from any point outside a circle to that circle always have equal length.

AP=AS,\quad BP=BQ,\quad CQ=CR,\quad DR=DS

Detailed analysis

Let $P,Q,R,S$ be the points where $\Gamma$ touches $AB,BC,CD,DA$ respectively. The standard equal-tangent-length property of a circle inscribed in a quadrilateral gives $AP=AS$ , $BP=BQ$ , $CQ=CR$ , $DR=DS$ .