Show that $N$ is orthcenter of triangle $AYZ$

Question

Triangle $ABC$ has height $BE,CF$.$M$ is the midpoint of $BC$. $EF$ and AM intersects at point $N$. Draw $NX\perp BC, XY\perp AB, XZ\perp AC(X\in BC, Y\in AB, Z\in AC)$.Show that $N$ is orthcenter of triangle $AYZ$.

I can't find any lemma to solve this problem. How we draw any another geometry element to solve.Show me pls!

Can you attach a picture? – Love Invariants Dec 08 '18 at 12:34 — Love Invariants, Dec 08 '18 at 12:34
I have added the picture. Pls show me this problem – Trong Tuan Dec 09 '18 at 01:47 — Trong Tuan, Dec 09 '18 at 01:47
What is the source of this problem? – Carl Schildkraut Dec 09 '18 at 03:09 — Carl Schildkraut, Dec 09 '18 at 03:09
Help me pls. I don't know where it is – Trong Tuan Dec 09 '18 at 03:42 — Trong Tuan, Dec 09 '18 at 03:42

score 3 · Answer 1 · answered Dec 09 '18 at 04:16

First note that $\triangle AEF$ and $\triangle ABC$ are similar. Let $P$ be the midpoint of $EF$, then $$\angle APE = \angle AMB\tag{1}.$$ Moreover, $E,F$ lie on the circle with center $M$ and diameter $BC$, so $MP\perp EF$. So $MNPX$ is circular. It follows that $$\angle NPX + \angle AMB = 180^\circ.\tag{2}$$ From (1) and (2) we conclude that $A,P,X$ are colinear. Consequently, $$\frac{BX}{XC} = \frac{NE}{NF}.\tag{3}$$ Since $XZ\parallel BE$, we have $EZ : ZC = BX:XC = NE:NF$. Thus $NZ\parallel CF$, and similarly $NY \parallel BE$ and we are done.

Note: (3) may be a little subtle but I leave it to you to clarify.

Your solution is nice. But it lacks proper explanation. Might be not so good for newbies. You jumped to so many conclusions. — Love Invariants, Dec 09 '18 at 18:14

Show that $N$ is orthcenter of triangle $AYZ$

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