Angle between two midpoints equals angle between point at perpendicular from top to base to other midpoint

Let, X, Y, Z be the midpoints of the sides AB, BC, CA of the triangle ABC. Let P be defined on BC so that ∠CPZ = ∠YXZ. Prove that AP is perpendicular to BC.

This question is from a book I'm reading about geometry (no answers) and I have been stuck on it for days - sorry if I'm just being stupid...

• You must show your own efforts to get your question answered. – Harish Chandra Rajpoot Jun 1 at 22:23
• Hint: Will Z lie on the perpendicular bisector of CP? – Mick Jun 2 at 4:51

If you have followed the hint posted by $$\mathbf{Mick}$$, you could have already solved this problem. The answer described below does not use that hint.
In order to make explaining easy, we need to draw a line to join $$Y$$ and $$Z$$. Furthaermore, let $$\measuredangle ZPC=\phi$$ and $$\measuredangle APZ=\alpha$$. We are also aware that $$\measuredangle ZXY = \measuredangle ZPC=\phi$$.
Since $$X$$, $$Y$$, and $$Z$$ are the midpoints of the three sides of the triangle $$ABC$$, the two triangles $$CZY$$ and $$XYZ$$ are directly similar. As a consequence, we have, $$\measuredangle YCZ = \measuredangle ZXY=\phi.$$
Since $$\measuredangle PCZ = \measuredangle ZPC$$, the triangle $$PCZ$$ is an isosceles triangle. Therefore, $$ZP=CZ$$. We also know that $$ZA=CZ$$, because $$Z$$ is the mid point of $$CA$$. This makes $$ZA=PZ$$ and the triangle $$APZ$$ an isosceles triangle. That means $$\measuredangle ZAP = \measuredangle APZ = \alpha$$.
Now consider the sum of the three vertex angles of the triangle $$APC$$. $$\measuredangle APC+\measuredangle PCA+\measuredangle CAP=\measuredangle APZ+\measuredangle ZPC +\measuredangle PCA+\measuredangle CAP=\alpha+\phi+\phi+\alpha=2\left(\alpha+\phi\right)=\pi$$ $$\therefore\quad \alpha+\phi=\frac{\pi}{2}$$
Now we have, $$\measuredangle APC=\measuredangle APZ+\measuredangle ZPC=\alpha+\phi=\frac{\pi}{2}.$$