Geometry proof involving triangle Let $ABC$ be an acute angle triangle with $\angle B>\angle C$. Let
$M$ be mid point of $BC$. Points $E$,$F$ are feet of altitudes from $B$ and $C$. Points $K$,$L$ are mid points of segments $ME$ and $MF$. let $T$ be point of $KL$ such that $TA$ is parallel to $BC$. 
Prove that $TA=TM$.
I have tried to solve it but was unable to even start it. Ding by coordinate geometry is too messy. Any help is appreciated.  
 A: 
Few elementary things: 


*

*Quadrilaterals $EHFA$ and $BCFE$ are cyclic.

*$MB=ME=MF=MC$
First we have to prove an important fact:

Segments $ME,MF,TA$ are tangent to circle $EHFA$

Let's prove it for segment $MF$. We'll use the fact that the angle between the tangent and the chord must be equal to the inscribed angle constructed above the same chord. In other words, if $MF$ is tangent to the circle $EHFA$ then:
$$\angle MFE=\angle FAE=\alpha\tag{1}$$
We can easily prove that (1) is true:
$$\angle MFE=\angle BFE+\angle MFB=\angle BCE+\angle MBF=(90^\circ-\beta)+(90^\circ-\gamma)=(180^\circ-\beta-\gamma)=\alpha$$
You can prove that $ME$ and $TA$ are tangent to circle $EHFA$ in exactly the same way. 
Back to the original problem. 
Consider two circles: 


*

*The first one is circle $EHFA$ 

*The second one is circle with center $M$ and radius equal to zero (yes, it sounds funny but circles with zero radius are actually quite useful in geometry). 


Segements $ME$ and $MF$ are tangent to both circles. Radical axis of these two circles has to pass through midpoints of both common tangents. In other words, line $KL$ represents radical axis of the two circles and the point $T$ is on the radical axis. 
Consequentially, the point T has the same power with respect to both circles:
$$Pow_{EHFA}(T)=Pow_M(T)$$ 
$$TA^2=TM^2$$ 
$$TA=TM$$ 
