CMI Entrance Exam 2019 #4

Let $ABCD$ be a parallelogram. Let $O$ be a point in its interior such that $\angle AOB+\angle DOC=180^\circ$. Show that $\angle ODC = \angle OBC$. 



We translate $O$ to $O'$ and $D, C$ to $A, B$, respectively. Thus $AO'OD, O'BCO$ are parallelograms. Now $\angle AO'B = \angle DOC$ because translations preserve angles, and it's given that $\angle DOC + \angle AOB = 180^{\circ}$, so $AO'BO$ is cyclic. This means $\angle O'AB = \angle O'OB = \angle OBC$ where the last equality results from parallelogram $O'OCB.$ Thus we are done. $\square$

Comments

Popular posts from this blog

1995 IMO #2

Inequality problem I made