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...

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

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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, $$\measuredangle APC=\measuredangle APZ+\measuredangle ZPC=\alpha+\phi=\frac{\pi}{2}.$$

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