Let $ABC$ be triangle with $AA_1$ , $BB_1$ and $CC_1$ as its altitude. $M$ and $N$ are the projection of $C_1$ to $AC$ , $BC$ respectively. Let $MN$ intersects $B_1C_1$ at $P$. Show that $P$ is midpoint of $B_1C_1$.
I have tried using Menelaus’ Theorem and similar triangles. Then , I use trigonometry and complex bashing. But , still can’t prove it. Should I draw some more line(s) or use any theorem? Can anyone give me some hints (or solution) please. Thank you!
Let $MN$ cut $BB_1$ at X.
$\angle CMC_1 = \angle CNC_1 = 90^0$ means $CMC_1N$ is cyclic. Then, $\angle NMC_1 = \angle NCC_1$.
Note also that $\angle C_1B_1 B = \angle NCC_1$ because of the properties of orthic triangle.
This means $MB_1XC_1$ is cyclic and is a rectangle. Result follows.