Prove that $\angle FME=90$ (WITHOUT using triangle identity) and show that $ME=MF$

Question

Prove that $\angle FME=90$ ( DO NOT use triangle identity) and show that $ME=MF$ (Triangle $ABC$ is right at $A$ and $AB=AC$ and $M$ is at the middle of $BC$,also $D$ is an arbitrary point on $BC$ and $DE$ and $DF$ are perpendicular to $AB$ and $AC$ respectively.)

I have already solved this using triangle identity technique(triangles $AMF,BME$ ) but don't know other method to solve it! Though was not able to prove $\angle FME=90$!

Answer 1

The four points $F,D,E$ and $A$ are concyclic. Also, since $\angle{AMD}+\angle{DEA}=180^\circ$, the four points $M,D,E$ and $A$ are concyclic.

Therefore, the five points $F,D,E,A$ and $M$ are concyclic.

It follows from this that $$\angle{EFM}=\angle{EAM}=45^\circ\quad\text{and}\quad \angle{FME}=\angle{FDE}=90^\circ$$

Answer 2

Hint One way is to look at the triangle $ABC$ and its $90^{\circ}$ clock-wise rotation around point $M$. Show that $F$ is the image of $E$.

Or more directly: Look at triangles $FMA$ and $EMB$. Anything special about them? That should do it.

Answer 3

It is quite trivial with coordinate geometry. Assume that $AE=x$ and $AF=y$. Then $$ D=(x,y),\quad B=(x+y,0),\quad C=(0,x+y),\quad M=\left(\frac{x+y}{2},\frac{x+y}{2}\right) $$ and

$$ MF^2 = \left(\frac{x+y}{2}\right)^2+\left(\frac{x-y}{2}\right)^2 = ME^2 = \frac{1}{2} EF^2$$ so $EMF$ is an isosceles right triangle by the Pythagorean theorem.