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 3 of 4: Match two right triangles using the cyclic angle condition

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In plain words

Two right triangles with the same short leg and the same acute angle are forced to be identical in shape and size — like two identical set-squares, one just rotated into place.

\triangle OLX \cong \triangle OMC \ \Longrightarrow\ LX = MC

Detailed analysis

Both $\triangle OLX$ and $\triangle OMC$ are right triangles at $L$ and $M$ (radius meets tangent at $90^\circ$ ), with equal legs $OL=OM$ (radii). Since $AX=AO$ , triangle $OAX$ is isosceles, giving $\angle OXL = 90^\circ-\tfrac{\angle A}{2}=\tfrac{180^\circ-\angle A}{2}$ ; because $ABCD$ is cyclic, $\angle A+\angle C=180^\circ$ , so this equals $\tfrac{\angle C}{2}$ . On the other side, $OC$ bisects $\angle C$ because $CM,CN$ are the two tangents from $C$ , so $\angle OCM=\tfrac{\angle C}{2}$ too. Matching angle and leg makes $\triangle OLX\cong\triangle OMC$ , hence $LX=MC$ .