Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Prove it Soviet Union

Problem:
A circle center $O$ is inscribed in $ABCD$ (touching every side). Prove that $\angle AOB + \angle COD$ equals $180$ degrees.

Solution

Solution:
Let $AB$ touch the circle at $W$ , $BC$ at $X$ , $CD$ at $Y$ , and $DA$ at $Z$ . Then $AO$ bisects angle $ZOW$ and $BO$ bisects angle $XOW$ . So $\angle AOB$ is half angle $ZOX$ . Similarly $\angle COD$ is half angle $XOZ$ and hence $\angle AOB + \angle COD$ equals $180$ .

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