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.
(Someone draw a picture of this because I don't know how)
To answer your second question, yes, a segment between two midpoints is parallel to the triangle's opposite side. In this case, call the intersection between $\overline{BH}$ and $\overline{MN}$ $P$, and draw altitude $\overline{MH'}$ perpendicular to $\overline{AC}$. Notice that $\overline{MH'} || \overline{BH}$, so $\angle AMH' = \angle ABH$, and therefore, $\angle BAC = \angle BMN$, so $\overline{MN} || \overline{AC}$.
Additionally, in this case $\overline{MN}$ bisects $\overline{BH}$. This is because $\triangle AMH' \cong \triangle MBP$ via Angle-Side-Angle, so $MH' = BP$. We also have $MH' = PH$, so $\overline{MN}$ does indeed bisect $\overline{BH}$, and it forms right angles because $\overline{MN} || \overline{AC}$.
Best Answer
This follows from the Intercept Theorem .
Let $DE$ be the line segment joining the midpoint $D$ of $AB$ and the midpoint $E$ of $BC$. Draw a line parallel to to $DE$ that passes through $A$. Extend the side $BC$ so that it intersects this line in the point $F$. By the intercept theorem, $$ {DB\over DA}={BE\over EF} $$
But $DB=DA$, so, $EF=BE=EC$. It follows that $F=C$, and, thus, $AC$ is parallel to $DE$.
(Of course, you could argue using similar triangles too. The intercept Theorem is equivalent to "the similar triangle business".)