Orthocenter and midpoint proof Let $H$ be the orthocenter of a triangle $\Delta ABC$. Let $M$ be the midpoint of $BC$, and let $E$, $F$ be the feet of the $B$ and the $C$ altitudes onto the opposite sides. Let $X$ be the intersection of ray $MA$ with the circumcircle of $BHC$. 
Prove that 
(1) $HX$, $EF$ and $BC$ concur at a point, say $P$;
(2) line $MX$ is perpendicular to the line $XH$, where $O$ is the center of the circumcircle; and
(3) also show that the line joining that point $P$ and $A$ is perpendicular to the line $HM$.
 A: Let $AD$ be the third altitude of $\triangle ABC$. Then, $DEF$ forms the pedal triangle which has the properties of having $H$ being its in-center. Therefore, $\theta = \theta_1$, for example. {Note: This part may not be necessary but I will leave it on for reference purpose.}
Extend $AD$ to cut the circle $BHC$ at $S$. Join $BS$ and $CS$. Then, $\alpha = \alpha_1 = \alpha_2$.

Added:  By the properties of the pedal $\triangle DEF$, $\alpha_3 = \alpha_4$.

WRT circle AFHE, $\alpha_3 = \alpha$.
WRT circle HDCE, $\alpha_4 = \alpha_1$.
WRT circle HBSC, $\alpha_1 = \alpha_2$.
This makes $BACS$ a kite.


Therefore, $\omega = \omega_1$.
Construct $BT$//$AC$ cutting the circle $BHC$ at $T$. Join $CT$. All dark green shaded angles are equal. In addition, eventually, $\omega_1 = \omega_2$ (through subtended arcs). This means $BCTS$ is an isosceles trapezium with $BC$//$ST$. This proves $ABTC$ is a parallelogram with $AMT$ is a diagonal (i.e. $AMT$ is in fact a straight line.)
$DHXM$ is then cyclic because the purple shaded angles are all equal. This completes the proof of (2).
Then, $H$ is the orthocenter of $\triangle APM$. This completes the proof of (3).
For (1), we can argue in the following way:
a) Let $FE$ cut $AM$ at $U$. b) Draw $UV \bot PM$ and let it cut $PX$ at $W$.
So far we have fixed (i) two sides ($MP$ and $MU$); (ii) three points ($U$, $M$ and $P$); (iii) two possible altitudes ($UV$ and $PX$); and (iv) the possible orthocenter ($W$).
It is sufficient to claim (I) $PU$ is the third side of the triangle $MPU$; (II) For $\triangle MPU$, $PU$ and $PM$ are two of the adjacent legs with $PX$ being the included altitude; and (III) The three lines are concurrent at $P$.
