Problem 1

A circle has its center on the side

AB

of the cyclic quadrilateral

ABCD

. The other three sides are tangent to the circle. Prove that

AD + BC = AB

Step 2 of 4: Build an auxiliary point on line

AD

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In plain words

This is the classic olympiad trick of 'copying' a length onto a more convenient line so that two separated pieces of the figure can finally be compared side by side.

AX = AO,\qquad X\in\text{line } AD

Detailed analysis

Since $O$ lies on $AB$ , mark a point $X$ on line $AD$ , on the same side of $A$ as $D$ , with $AX=AO$ . This turns the length $AO$ (which sits on side $AB$ ) into a length measured along side $AD$ , where the tangent point $L$ already lives — setting up a direct comparison with triangle $OMC$ .