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 1 of 4: Name the points of tangency

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In plain words

Whenever two lines from a single outside point just graze a circle, they must reach it at exactly the same distance — think of two ropes of equal length swung from a peg until they both become tangent to a circular fence.

CM = CN \quad\text{and}\quad DM = DL

Detailed analysis

Let the circle touch sides $AD$ , $CD$ , $BC$ at $L$ , $M$ , $N$ respectively. Since two tangent segments drawn from the same external point to a circle are equal in length, the tangents from $C$ satisfy $CM=CN$ , and the tangents from $D$ satisfy $DM=DL$ .