Symmedian and orthic triangle properties I am trying to prove the following lemma that may be useful for junior-level international contests:

Given $\Delta ABC$ an acute triangle and $(BE)$ and $(CF)$ its altitudes. Let's consider $(AM)$ the median line, $M \in (BC)$ and $AM \cap EF = \{N\}$. If $NP \perp BC$, $P \in (BC)$, then $AP$ is the $A$-symmedian in the triangle $\Delta ABC$.

Attempt. I have tried using the following metric relations in a triangle:

*

*$s_a = \frac{2bc}{b^2 + c^2} m_a$

*$m_a^2 = \frac{2(b^2 + c^2 - a^2)}{4}$ (Stewart)

*$\frac{BD}{DC} = {AB^2}{AC^2}$ if and only if $AD$ symmedian (Steiner)

Drawing $AD$ height, I proved that $\frac{PM}{DM} = \frac{a^2}{b^2 + c^2}$. I could not prove that $AP$ symmedian, although I have tried to calculate ratios and lengths using Pythagorean Theorem or trigonometric relations.
 A: Let $D$ be the mid-point of $EF$

Since $MF=ME=BC/2$, it follows that $MD\perp EF$. So $MNDP$ is a cyclic quadrilateral, and
$$\angle NDP = 180^\circ - \angle NMP = 180^\circ - \angle ADE.$$
Thus $A, D, P$ are colinear and it follows that $AP$ is the symmedian.
A: The two triangles $ABC$ and $AEF$ are inversely similar. This can be proven easily by first showing the quadrilateral $BCEF$ is cyclic.
We know that the corresponding lines of similar or inversely similar  triangles make equal angles. The sides $AB$, $BC$, and $CA$ of $\triangle ABC$ correspond to the sides $AE$, $EF$, and $FA$ of $\triangle AEF$ respectively.
Since $D$ is the midpoint of $EF$, $AD$ is the $A$-median of $\triangle AEF$. It is given that $AM$ is the $A$-median of $\triangle ABC$. Due to the inverse similarity of the two mentioned triangles,
$$\measuredangle DAF = \measuredangle MAC.$$
This make the line $AP$ the isogonal conjugate of the $A$-median of $\triangle ABC$. This means that, by definition, $AP$ is $A$-symmedian of $\triangle ABC$.
