Triangles $ABH$ and $ABK$ are right triangles whose hypothenuse is $AB$. So they have the same circumcircle, whose diameter is $AB$. As a consequence, considering that $M$ is the midpoint of the diameter $AB$, we get that $MH=MK$ because they are radii of the same circumference, and then $MHK$ is isosceles.
Without using circles: we can show that, in any right triangle, the median drawn to the hypotenuse is equal to half the hypotenuse. To show it for the right triangle $AHB$,
let us draw a line $ME$ starting from the midpoint $M$, parallel to the
leg $AH$, and ending to the intersection point $E$ with the other leg $HB$. We know that the angle $AHB$ is right. The angles $MEB$ and
$AHB$ are congruent, since they are corresponding angles if we consider the parallel lines $AH$ and $ME$, and the transverse line $HB$. So, the angle $MEB$ is a right angle.
Because the line $ME$ starts from the midpoint $M$ and is parallel to $AH$, it divides the leg $HB$ in two congruent segments $HE$ and $EB$ with equal length. These considerations show that the triangles $MEH$ and $MEB$ are right triangles, have congruent legs $HE$ and $EB$, and a common leg $ME$. This means that these triangles are congruent. We then get that the segments $MH$ and $MB$ are also congruent, since they are corresponding sides (hypothenuses) of these triangles. So we have shown that, in the right triangle $AHB$, the median $MH$ is equal to half the hypotenuse $AB$.
Applying the same procedure to the right triangle $BKA$, we get that $MK$ is congruent with $MA$, so that it also equals half of the hypothenuse $AB$.
We then conclude that $MH=MK$, and so the triangle $MHK$ is isosceles.
You can use
$$ \dfrac{[ADE]}{[ABC]} = \dfrac{\tfrac{1}{2}\cdot AD \cdot AE \cdot \sin A}{\tfrac{1}{2}\cdot AB \cdot AC \cdot \sin A} $$
to know their ratios. And you know their difference.
Best Answer
Let $P$ be the point on $AC$ such that $FP$ is perpendicular to $AC$.
Let $Q$ be the point on $BC$ such that $FQ$ is perpendicular to $BC$.
Let $R$ be the point on $DE$ such that $FR$ is perpendicular to $DE$.
Since $FD$ is the angle bisector of $\angle ADE$, we have $|FP| = |FR|$. Similarly $|FQ| = |FR|$.
Thus we have $|FP| = |FQ|$. Also $\angle A = \angle B$ and $\angle APF = 90^\circ = \angle BQF$, therefore the two triangles $\triangle APF$ and $\triangle BQF$ are congruent, which implies that $|AF| = |BF|$.