proving angle formed by simson line equals another angle

Question

P is a point on the circumcircle of $\triangle{ABC}$ and $O$ is its circumcentre. Prove that $\angle{APO} =$ the angle between the Simson line of $P$ and $BC$.

Here's a diagram I drew on geogebra of the problem, where $NLM$ is the Simson line. I'm trying to prove the red angles are equal (the 22.8 is arbritrary, just for me to mark the angles.) I've tried creating an isoceles triangle by constructing $OA$, but I'm really not sure how to angle chase this. I think I might need to use subtended angles with cyclic quadrilaterals, or maybe some constructions? I'm not able to connect the dots though. Could someone lead to the right direction?

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Answer 1

Hint:

1. $\angle ACP$ is half of $\angle AOP$;
2. $\angle ACP$ is thus directly related to $\angle APO$;
3. $\angle ACP$ is also directly related to $\angle MLC$ via an inscribed quadrilateral.

Answer 2

Let's assume that $\angle APO=\angle OAP=x$. Then,

$$\angle MAP=x+\angle MAO=x+90^{\circ}-\angle B \implies \angle APM=\angle B-x.$$

On the other hand,

$$\angle B=\angle APC=\angle APM+\angle MPC \implies \angle MPC=x.$$

But $MCPL$ is cyclic. So, $\angle MLC=\angle MPC=x.$

We are done.