X lies on Polar of T. Let $ABC$ be a triangle and $D,E,F$ are the foot of altitudes from $A,B,C$ on $BC,AC,AB$ respectively and $H$ be the orthocenter of $\triangle ABC$, $M$ be the midpoint of $BC$, let $MH\cap EF=X$ and $EF\cap BC=T$. Prove that $X\in$ Polar of $T$ WRT $\odot(ABC)$
What I did: project from H from line EF to BC
 A: Configurations involving a triangle $\Delta{ABC}$, its circumcircle $(ABC)$, orthic triangle $\Delta{DEF}$, and line $MH$ are rich with concurrences, collinearities and cyclics.  There's quite a bit going on, and I'd recommend looking at

*

*blog post The Ex-points and the Queue-points - Part One

*Pamfilos' pages Orthocenter and Orthocenter 2.

Much of what I'll say is demonstrated in the above references.
Here is the diagram:

The circles $(BCEF)$ and $(AFHE)$ are shown, and $Q$ is the second intersection of $(AFHE)$ and $(ABC)$.  $O$ is the center of $(ABC)$ and $AA'$ is a diameter.  The references show that $Q,H,M,A'$ are collinear.  $QA$ runs through $T$ because $T$ is the radical center of the three circles. Finally, $P=AX\cap A'T$.
Note that in triangle $\Delta ABC$ the points $A$ and $H$ are on each other's polars wrt $(BCEF)$ because $\Delta THA$ is the diagonal triangle of quadrangle $BCEF$. (The altitudes $AD,BE,CF$ are some of the lines we would use in the straightedge construction of $H$ and $A$'s polars).  Cutting to the chase, we claim that $X=EF\cap MH$ is the orthocenter of the triangle $\Delta AA'T$.  If this is the case, and $XQ$ and $XP$ are altitudes, $T$ and $X$ are on each other's polars wrt to the circle $(A'AQP)$.
Line $XM\perp QA$ because $\angle{HQA}=\angle{HFA}=\pi/2$.  Line $XE\perp AA'$ because, respectively, the diameters through $A,B,C$ are perpendicular to the tangents through $A,B,C$, which are parallel to the sides $EF,FD,DE$ (see this answer).  Thus $X$ is the orthocenter of triangle $\Delta{TAA'}$ and $XP\perp A'T$.  Since $\angle{APA'}=\pi/2$, $P$ lies on the circle $(ABC)$.
Since $T$ and $X$ are on each other's polars wrt to the circle $(A'AQP)$, and circles $(A'AQP)$ and $(ABC)$ are the same, we are done.
