Points $O$ and $H$ are the circumcenter and orthocenter of acute triangle $ABC$, respectively. The perpendicular bisector of segment $AH$ meets sides $AB$ and $AC$ at $D$ and $E$, respectively. Prove that $\angle DOA= \angle EOA$.
After drawing the diagram; I was stuck
so, I figured I knew the coordinates of all points so I should coord bash this
but it was a very tedious computation and I must have made a mistake somewhere as I couldn't get the right solution. (P.S. if you want me to share my failed coordinate bash, do let me know though my handwriting is not very good).
A elementary solution/trigonometrical/coordinate bash would be ideal but any solution would mean a lot.
Thanks!
Best Answer
First see that $\triangle ADH\sim\triangle AOC$, so $\triangle ADO\sim\triangle AHC$ (either by spiral similarity, or look at length ratios).
Similarly, $\triangle AEO\sim\triangle AHB$. So $\angle DOA=\angle HCA=\angle HBA=\angle EOA$.