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 4 of 4: Combine both sides to finish

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In plain words

Once both halves of $AB$ have been rewritten in terms of the same four tangent pieces that make up $AD$ and $BC$ , the two sides of the equation are literally built from identical building blocks.

AB = (AL+MC)+(BN+MD) = AD+BC

Detailed analysis

From $AX=AL+LX$ and $LX=MC$ we get $AO=AL+MC$ . A symmetric construction at vertex $B$ (swap $A\leftrightarrow B$ , $D\leftrightarrow C$ ) gives $BO=BN+MD$ . Adding, $AB=AO+OB=AL+MC+BN+MD$ . Using the equal-tangent facts $MC=CN$ and $MD=DL$ from Step 1, this becomes $AB=(AL+LD)+(BN+NC)=AD+BC$ .