In the following figure prove that: $ME=MF$ 
$ABC$ is a right isosceles triangle($\angle A=90 , AB=AC$).We draw two perpendicular lines from an arbitrary point $D$ on side $BC$ to the other sides.If $M$ is the midpoint of $BC$ prove that: $ME=MF$


I think it's a challenge to show: $\angle MEF=\angle MFE$!!
 A: $\angle \, DMA = \angle \, DEA = 90^{\circ}$ so $DMEA$ lie on a common circle. This means that point $M$ lies on the circumcircle of triangle $DEA$. However, $F$ also lies on that circle, because $DEAF$ is a rectangle. Hence, $$\angle \, MEF = \angle \, MAF = 45^{\circ}$$
$$\angle \, MFE = \angle \, MAE = 45^{\circ}$$ Thus triangles $EMF$ is isosceles with $\angle \, MEF = \angle \, MFE = 45^{\circ}$ and therefore $ME=MF$ as well as $ME$ is orthogonal to $MF$.
A: I like @Futurologist's solution much better, but here's a reasonably-self-evident picture proof.

A: Construct the following points on a coordinate plane, let $A(0,0)$, $B(0,y)$, $C(x,0)$ and $D(a,b)$. Then $M(x/2,y/2)$, $E(0,b)$ and $F(a,0)$.  Now $$ME^2=(x/2)^2+(y/2-b)^2=(y/2)^2+(x/2)^2+b^2-by$$ and $$MF^2=(y/2)^2+(x/2-a)^2=(y/2)^2+(x/2)^2+a^2-ax.$$ We also know that $D$ lies on $BC$, therefore the slope of line $BD$ and $DC$ is the same. This implies $$\frac{b-y}{a}=\frac{b}{a-x}\Rightarrow ay+bx=xy.$$ Now since $x=y$ we have: $\Rightarrow a+b=x\Rightarrow b^2-a^2=x(b-a)\Rightarrow b^2-xb=a^2-ax\Rightarrow b^2-by=a^2-ax.$ Thus, $ME=MF.$
