If $P$ is on the circumcircle of a triangle, show that the feet of the perpendiculars from $P$ to the side-lines of the triangle are collienar

Question

Let $ABC$ be a triangle and $P$ be any point on its circumcircle. Let $X,Y,Z$ be the feet of the perpendiculars from $P$ onto lines $BC, CA$ and $AB$. Prove that points $X,Y,Z$ are collinear.

So I've already made a diagram(it is attached below), but I don't know how to prove it from there. Please help and explain your solution thoroughly because I have a test about this tomorrow and I want to understand this! Thank you! :D

Answer 1

The method of proof is to show that $\displaystyle \angle NMP+\angle PML=180^{\circ }$.

$\displaystyle PCAB$ is a cyclic quadrilateral, so $\displaystyle \angle PBA+\angle ACP=\angle PBN+\angle ACP=180^{\circ }$.

$\displaystyle PMNB$ is a cyclic quadrilateral (Thales' theorem), so $\displaystyle \angle PBN+\angle NMP=180^{\circ }$.

Hence $\displaystyle \angle NMP=\angle ACP$. Now $\displaystyle PLCM$ is cyclic, so $\displaystyle \angle PML=\angle PCL=180^{\circ }-\angle ACP$.

Therefore $\displaystyle \angle NMP+\angle PML=\angle ACP+(180^{\circ }-\angle ACP)=180^{\circ }$.

Answer 2

Hint: Do you see some cyclic quads? there are 3 cyclic quads . Try to draw a neat diagram and you will observe the 3 cyclic quads.

Answer 3

Because $$\measuredangle AYX=\measuredangle APX=\measuredangle XPZ-\measuredangle APZ=180^{\circ}-\beta-\measuredangle APZ=$$ $$=\measuredangle APC-\measuredangle APZ=\measuredangle ZPC=\measuredangle ZYC.$$