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<AOB + < COD = \pi inside parallelogram ABCD => < CBO =< CDO

Source: Switzerland - Swiss TST 2002 p2

February 18, 2020
geometryparallelogramequal angles

Problem Statement

A pointO O inside a parallelogram ABCDABCD satisfies ∠AOB+∠COD=π\angle AOB + \angle COD = \pi. Prove that ∠CBO=∠CDO\angle CBO = \angle CDO.
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