$O$ is the orthocentre of $\triangle ABC$ if $AP\perp BC$, $BR\perp AC$ and $CQ \perp AB$. Prove that $\angle OPQ= \angle OPR$

Question

$O$ is the orthocentre of $\triangle ABC$ if $AP\perp BC$, $BR\perp AC$ and $CQ \perp AB$. Prove that $\angle OPQ= \angle OPR$

Answer 1

As $APQC$ is cyclic we get that $\angle BPQ = \angle BAC$. Similarly as $ABPR$ is cyclic we get that $\angle CPR = \angle BAC$. Then we have:

$$\angle APQ = \angle BPA - \angle BPQ = \frac{\pi}{2} - \angle BAC$$

$$\angle APR = \angle CPA - \angle CPR = \frac{\pi}{2} - \angle BAC$$

Thus we get $\angle APQ = \angle APR$. Hence the proof.