We can do it without drawing any line.
Let our triangle be $ABC$, with $\angle B=\angle C$.
By ASA, $\triangle ABC$ and $\triangle ACB$ are congruent.
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.
Best Answer
This is only true if $AB=AC$ (that is, if $\overline{BC}$ is the base of the isosceles triangle). If that's true, then $\angle ABC\cong\angle ACB$ by the Isosceles Triangle Theorem, and $\angle ACB\cong\angle A'CB$ because reflection preserves angle measure. You should then be able to use a theorem or postulate about parallel lines cut by a transversal to get that $\overline{AB}\parallel\overline{C'A'}$.