Denote the center of incircle by $O$ and of circumcircle by $O'$. It's easy to caclulate that
$$\angle ABC=\angle ACB= 2\angle OBC=2\arctan\frac{r}{BC/2}=2\arctan\frac{1}{2}$$
Thus we can calculate the height $h$ with base $BC$ is
$$h=\frac{BC}{2}\tan\angle ABC=24\tan(2\arctan\frac{1}{2})=24\times\frac{2\times\frac{1}{2}}{1-(\frac{1}{2})^2}=32$$
By symmetry, $O'$ shall lie on the height $h$. Consider the property of circumcircle that $O'A=O'B=O'C=R$
$$O'B^2=O'C^2=(\frac{BC}{2})^2+(h-R)^2=R^2$$
which gives the solution
$$R=25$$
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
From the figure
one imediately derives the equations $${\rho\over h}={a\over 2R}\ ,\qquad{\rm i.e.,}\qquad a h=600\ ,$$ $$h+\rho+d=2R\ ,\qquad{\rm i.e.,}\qquad h+d=38\ ,$$ $$2R d=a^2 \ ,\qquad{\rm i.e.,}\qquad 50d = a^2\ .$$ Eliminating $h$ and $d$ one obtains a cubic equation for $a$, two of whose solutions are natural numbers. Given $a$ and $d$ one computes $s=\sqrt{4R^2-a^2}$, and the base $b$ of the triangle is $b=2\sqrt{a^2-d^2}$.