To find the measure of the given angle 
In triangle $ABC$ angle $\widehat C=60°$. $(AD)$ and $(BE)$ are perpendicular on $(BC)$ and $(AC)$ respectively. $M$ is the midpoint of $[AB]$. How to find the measure of angle $\widehat{EMD}$ in degrees?.
 A: hint:
$MD=ME=MB=MA$(why?)
$\angle MBD=\angle MDB,\angle MAE=\angle MEA$
can you go from here?
Edit: I add details.
in a right triangle, the middle line of hypotenuse is half length of the hypotenuse. In Right Triangle $ABD,MD=MA=MB$, In Right Triangle $ABE,ME=MA=MB$
$\angle MEC=\pi-\angle AEM=\pi-\angle CAB, \angle MDC=\pi-\angle MDB=\pi-ABC$
$\angle ABC+\angle CAB=\pi-\angle ACB,\angle EMD+ \angle MEC+\angle MDC+\angle ACB=2\pi$
$\angle EMD=\pi-2\angle ACB$
A: Here is a heavyweight, straightforward solution.
Putting: $AB=c$, $AC=b$ and $BC=a$.
Let us see $\triangle DMB$ first:


*

*$BM=\frac{c}{2}$

*$BD=c\cos B$ [why?]


So, applying the cosine rule on $DMB$, we have $DM=\sqrt{(\frac{c}{2})^2+c^2\cos ^2B-c^2\cos ^2B}=\frac{c}{2}$. Similarly, doing this on $\triangle MAE$, yields $ME=\frac{c}{2}$.  
On $\triangle DCE$:


*

*$CD=b\cos C$ [why?]

*$CE=a\cos C$   


So, applying the cosine rule on $DCE$, we have:
$$DE=\sqrt{b^2\cos ^2C +a^2\cos ^2C-2ab\cos ^3C}=\cos C\sqrt{b^2 +a^2-2ab\cos C}=c\cos C=\frac{c}{2}$$ So, $ME=DM=DE=\frac{c}{2}$. Thus, $MED$ is an equilateral triangle. 
This proof tells us more about the question. We have shown that $DMB$ is isosceles for ever acute triangle, not necessarily when $\angle C$ is $60^{\circ}$. However, it is equilateral if and only if $\angle C= 60^{\circ}$, since $\cos C = \frac{1}{2}$, only when $\angle C= 60^{\circ}$, when $C$ is an angle of a triangle .     
