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've made a big deal out of it.
Complete the circle from the quadrant for a better understanding.
The smaller circle touches the larger circle at the point $P$. Draw a tangent to the smaller circle at the point $P$. Since the larger circle touches the smaller circle at $P$ as well, the same tangent is also a tangent to the larger circle.
Now, recall:
the perpendicular drawn from a tangential line at the point of tangency passes through the centre of the circle
Therefore, if we draw a perpendicular from the tangent at $P$, it will pass through the center $O$ as well as the center $B$ as it is a tangent to both the circles. Since the perpendicular at $P$ passes through $O$ and $B$ both, it is obvious that the line $PB$ passes through $O$.
Best Answer
Let $M$ be the midpoint of the $EF$-side of the orthic triangle. It is enough to show that $MFB$ and $EHG$ are similar. It is simple to prove that $\widehat{BFM}=\widehat{BFE}=\widehat{GHE}=\widehat{DHE}=\pi-C$ by angle chasing. By the properties of the orthic triangle ($AEF$ and $ABC$ are similar) $FE=a\cos A$, while $FB=a\cos B$. It follows that $$\frac{FM}{FB}=\frac{a\cos A}{2a\cos B}=\frac{\cos A}{2\cos B}.$$ The distances of the orthocenter $H$ from the $AC$ and $BC$-sides are given by $$ HD = BD\cot C = c \cos B \cot C = 2 R \cos C \cos B $$ $$ HE = AE\cot C = c \cos A \cot C = 2 R \cos C \cos A $$ hence $\frac{FM}{FB}=\frac{HE}{HG}$ and we are done.