Question 26
The point $O$ is situated inside the parallelogram $ABCD$ so that
$$
\angle AOB + \angle COD = 180^\circ
$$
Prove that $\angle OBC = \angle ODC$.
$$
\angle AOB + \angle COD = 180^\circ
$$
Prove that $\angle OBC = \angle ODC$.
MathNet 1997 — Part 01 Question 26
Math Olympiad Vault · MathNet 1997 — Part 01