socrates wrote:

Given is an acute triangle $ABC$ and the median $AM.$ Draw $BH\perp AC.$ The line which goes through $A$ and is perpendicular to $AM$ intersects $BH$ at $E.$ On the opposite ray of the ray $AE$ choose $F$ such that $AE=AF.$ Prove that $CF\perp AB.$

I have changed the labelling a bit, to maintain familiarity.

Rephrased problem wrote:

Let $ABC$ be an acute triangle with orthocenter $H$ . Let $M$ be the midpoint of $BC$ . Rays $BH$ and $CH$ intersect the line through $A$ perpendicular to $AM$ at points $X$ and $Y$ , respectively. Prove that $AX = AY$ .

My Solution: Let $\overline{AH} \cap \overline{BC}=D, \overline{BH} \cap \overline{CA} = E, \overline{CH} \cap \overline{AB} = F, \overline{EF} \cap \overline{BC} = G$ and $\overline{AM} \cap \overline{GH} = K.$

First, we note that $BCEF$ is a cyclic quadrilateral, hence, by Brokard's $M$ is the orthocenter of $\triangle AHG.$ Thus, $\angle AKH = 90^{\circ}$ and $\overline{AM} \cap \overline{GH} = \{K\} \in \odot(AEHF)$ . Since, $\angle MAY = \angle MAX = 90^{\circ} \implies \overline{AY} \parallel \overline{HG}$ i.e. $\overline{XY} \parallel \overline{HG}-(*)$ thus, they intersect at $P_{\infty}$ , the point at infinity along $\overline{XY}.$

Let $A'$ denote the midpoint of $XY.$ Now, by Cevians Induce Harmonic Bundles Lemma $-1=(G,D;B,C)\stackrel{H} = (P_\infty,A;X,Y)$ But, by Midpoints and Parallel Lines Lemma $(X,Y;A',P_{\infty})=-1$ , thus $A'=A$ and we're done. $\quad \square$