In triangle $\Delta ABC$, let $M$ be the midpoint of $BC$. Denote the feet of perpendiculars from $C$ to $AB$ and $B$ to $AC$ by $D$ and $F$, respectively. Furthermore, let $H$ be the orthocenter of $\Delta ABC$ and $G$ denote the intersection of the lines $AH$ and $DF$. Extend $MG$ to meet the circumcircle of $\Delta ADF$---whose center $O$ is the midpoint of $AH$---at $E$. Show that $\measuredangle BEH=\measuredangle CEH$.
I thought why not solve a geometry problem out of nostalgia and curiosity whether I'd still remember the theorems. Turns out I do, but the solution is rather long and heavy. I'm only sketching the details and assuming familiarity with poles and polars and cross-ratios on lines and circles.
Step 1: Let $k$ be the circumcircle of $\triangle ADF$. We know that this circle $k$ is the circle with diameter $AH$, and passes through $E$. Thus, $EA\perp EH$.
Lemma 1. Let $u$, $v$, $p$ and $q$ be four distinct lines passing through one common point $P$. Assume that $p\perp q$. Then, $p$ and $q$ are the two angle bisectors (internal and external, in some order) of the angle between the lines $u$ and $v$ if and only if the lines $p$ and $q$ are harmonic with respect to $u$ and $v$.
This is a standard fact and shouldn't be difficult to prove. It is what allows one to use cross-ratios for verifying angle bisections.
We have to prove that $\measuredangle BEH = \measuredangle CEH$. In other words, we have to show that $EH$ is an angle bisector of the angle between the lines $BE$ and $CE$. By Lemma 1 (applied to $u=BE$, $v=CE$, $p=EA$, $q=EH$ and $P=A$), this boils down to proving that the lines $EA$ and $EH$ are harmonic with respect to $BE$ and $CE$.
Step 2: Recall the following fact:
Lemma 2. Let $m$ be a circle, and $X$, $Y$, $Z$, $W$ be four points on $m$. Then, the polar of the point $XZ\cap YW$ with respect to $m$ is the line connecting the points $XY\cap ZW$ and $YZ\cap WX$.
Applying Lemma 2 to $m=k$, $X=A$, $Y=F$, $Z=H$ and $W=D$, we conclude that the polar of the point $G$ with respect to the circle $k$ is the line connecting the points $C$ and $B$. In other words, the polar of the point $G$ with respect to the circle $k$ is the line $BC$.
Step 3: Let $E'$ be the point of intersection of the line $MG$ with the circle $k$ distinct from $E$. (Heuristical note: When a problem introduces a point of intersection of a line with a circle, always think about the other point. This is akin, and related, to studying a root of a polynomial using the other roots.) Then, $AH\cap EE'=G$. Now, Lemma 2, applied to $m=k$, $X=A$, $Y=E'$, $Z=H$ and $W=E$, yields that the polar of the point $G$ with respect to the circle $k$ is the line connecting the points $AE'\cap HE$ and $E'H\cap EA$. Since we already know that this polar is the line $BC$, this rewrites as follows: The line $BC$ is the line connecting the points $AE'\cap HE$ and $E'H\cap EA$. Hence, $AE'\cap HE\in BC$ and $E'H\cap EA\in BC$.
Step 4: Recall that we have to prove that the lines $EA$ and $EH$ are harmonic with respect to $BE$ and $CE$. By intersecting all lines with $BC$, this task transforms into proving that the points $EA\cap BC$ and $EH\cap BC$ are harmonic with respect to $B$ and $C$. But since $EA\cap BC = E'H\cap BC$ (this follows from $E'H\cap EA\in BC$), this transforms into proving that the points $E'H\cap BC$ and $EH\cap BC$ are harmonic with respect to $B$ and $C$. If we replace these points by the lines that connect them with $H$, we see that this is equivalent to proving that the lines $E'H$ and $EH$ are harmonic with respect to $BH$ and $CH$.
Step 5: A final lemma now:
Lemma 3. Let $m$ be a circle, and let $P$, $U$ and $V$ be three points on $m$. Let $\ell$ be a line through the pole of $UV$ with respect to $m$. Let $\ell$ intersect the circle $m$ in two points $S$ and $S'$. Then, the lines $S'P$ and $SP$ are harmonic with respect to $UP$ and $VP$.
This lemma isn't difficult to prove; you might recognize it as an equivalence between two of the various characterizations of harmonic quadrilaterals. Here is a quick proof of Lemma 3 using cross-ratios on circles (admittedly not the most well-known kind of cross-ratios, but too useful for me to avoid here):
Let $Q$ be the pole of $UV$ with respect to $m$. We know that $Q\in \ell$. We also know, by one of the basic properties of poles, that the lines $QU$ and $QV$ touch the circle $m$ at $U$ and $V$, respectively. In particular, the line $QU$ touches $m$ at $U$. A somewhat weird way to rewrite this fact is by saying that the second point of intersection of the line $QU$ with $m$ apart from $U$ is $U$ (since points of tangency naturally are viewed as double points of intersection). Now, let $Z$ be the point $\ell \cap UV$. By a well-known fact about polars, the points $Q$ and $Z$ are harmonic with respect to $S$ and $S'$. Equivalently, the points $S'$ and $S$ are harmonic with respect to $Q$ and $Z$. In other words, the lines $S'U$ and $SU$ are harmonic with respect to $QU$ and $ZU$ (here, we have just connected each point with $U$). Now, projecting this cross-ratio onto the circle $m$ from the point $U$, we conclude that the points $S'$ and $S$ are harmonic with respect to $U$ and $V$ on $m$ (since the second point of intersection of the lines $ZU$ with $m$ apart from $U$ is $V$, and the second point of intersection of the line $QU$ with $m$ apart from $U$ is $U$). Looking at this cross-ratio from the point $P$ (which lies on $m$), we rewrite this as follows: The lines $S'P$ and $SP$ are harmonic with respect to $UP$ and $VP$. Lemma 3 is proven.
Step 7: Let us now conclude the solution. We have $OD\perp MD$ (a consequence, for instance, of the fact that $OM$ is a diameter of the nine-point circle of $\triangle ABC$, whereas $D$ lies on this circle). Thus, $M$ lies on the tangent to $k$ at $D$. Similarly, $M$ lies on the tangent to $k$ at $F$. Altogether, $M$ is thus the point where the tangent to $k$ at $D$ intersects the tangent to $k$ at $F$. Thus, $M$ is the pole of $DF$ with respect to $k$. The line $MG$ passes through $M$ and intersects $k$ at $E$ and $E'$. Thus, Lemma 3 (applied to $m=k$, $P=H$, $U=D$, $V=E$ and $\ell = MG$) yields that the lines $E'H$ and $EH$ are harmonic with respect to $DH$ and $EH$. In other words, the lines $E'H$ and $EH$ are harmonic with respect to $CH$ and $BH$. In yet other words, the lines $E'H$ and $EH$ are harmonic with respect to $CH$ and $BH$. We are done.
Let $P$ be the pole of $MG$ with respect to $(O)$ (i.e, the circumcircle of $\Delta ADF$). $P$ must be the harmonic conjugate of $G$ with respect to $DF$, which implies that $P$ is in fact the intersection of $DF$ and $BC$. Since $P$ was the pole of $MG$ with respect to $(O)$ and $MG$ intersects $(O)$ at $E$, $PE$ must be tangent to $O$. Let's call this tangent $T$. The facts that $DFCB$ is cyclic and the lines $DF$,$BC$, and $T$ concur at $P$, imply that $PE$ is also tangent to the circumcircle of $\Delta EBC$ which we refer to as $(O')$. Consequently, there's a homothety with respect to $E$ maps $(O)$ to $(O')$. Now let $B'$ and $C'$ denote the intersection of $EB$ and $EC$ with $(O)$ , respectively. Then $B'C'$ is parallel to $BC$ and $AH$ is it perpendicular bisector. Thus $H$ is the midpoint of the arc $B'HC'$ in $(O)$ which is the desired result.
