1985
Problem - 3135
Let $O$ be a point inside a convex pentagon, as shown, such that $\angle{1} = \angle{2}, \angle{3} = \angle{4}, \angle{5} = \angle{6},$ and $\angle{7} = \angle{8}$. Show that either $\angle{9} = \angle{10}$ or $\angle{9} + \angle{10} = 180^\circ$ holds.
By the law of sine and given equal angle conditions:
\begin{align*}
\frac{OA}{\sin\angle{10}} &= \frac{OB}{\sin\angle{1}} = \frac{OB}{\sin\angle{2}}\\
&= \frac{OC}{\sin\angle{3}} = \frac{OC}{\sin\angle{4}}\\
&= \frac{OD}{\sin\angle{5}} = \frac{OC}{\sin\angle{6}}\\
&= \frac{OE}{\sin\angle{7}} = \frac{OE}{\sin\angle{8}}\\
&= \frac{OA}{\sin\angle{9}}
\end{align*}
Hence, $$\sin\angle{9} = \sin\angle{10} \implies \angle{9} = \angle{1}\quad\text{or}\quad\angle{9} +\angle{10}=180^\circ$$