Help me understand the answers

Question

THE PROBLEM

Let $ABCD$ be a rectangle and $E, F$ the means of the sides $AB$, respectively $AD.$ We know that there is a point $M$ on the segment $AC$ such that $\angle EMF = 60$ and $AC^2 = 4*ME*MF$ . Show that ABCD is square.

My idea

drawing

Okey, so the first thing i thought of is that drawing the heights $MX$ and $MY$ are going to help us.

What about writing the area of traingle $FEM$ as $ME**MF*\sqrt{3}/4$

Idk how the $AC^2 = 4*ME*MF$ ia going to help us.

If we show some congruences of angles and triangles in the rectangle $AXMF$ the problem is solved.

Answer 1

It all boils down to showing that $EMF$ is equilateral. To this aim, let $G$ a point such that $GF \cong EM$, and $\angle GFE \cong \angle EFM$.

Since $2EF \cong AC$, from $AC^2 \cong 4\cdot EM \cdot MF$ you get $EF^2 \cong EM\cdot MF$, i.e. $$EF:ME = MF:EF,$$ which means that $EMF \sim GEF$ (SAS criterion). In particular, $\angle GEF \cong \angle EFM$, since they are opposed to proportional sides. Hence the two triangles are isosceles. Since $\measuredangle EMF = 60^\circ$, we have, in fact, that they are equilateral.

EDIT (Prompted by OP's question in comment)

As a matter of fact, the hypothesis $\measuredangle EMF = 60^\circ$ is unecessary. The other hypothesis alone implies that $EMF$ is isosceles and hence that $AC$ perpendicularly bisects $EF$, which, in turn, yields $AE \cong AF$.

Answer 2

$2EF=BD=AC$ (midpoint theorem)

$EF^2=ME^2+MF^2-2\times ME\times MF \times \cos60^{\circ}$ (Cosine Theorem)

$EF^2=ME^2+MF^2- ME\times MF $

$ME^2+MF^2- 2\times ME \times MF=0$ $(AC=2EF, AC^2=4\times EF^2=4\times ME\times MF$ and $EF^2=ME\times MF)$

$(ME-MF)^2=0$

$ME=MF$

$\triangle AME\cong \triangle AMF$ (SSS congruency)

$\angle EAM=\angle FAM=45^{\circ}$

$\therefore\ ABCD$ is a square.