Consider the figure below:
Let's calculate $CD$ using the Law of cosines in $\triangle CAD$:
$$ CD^2=10^2+200^2-2 \cdot 10 \cdot 200 \cdot \cos 82^{\circ} \Rightarrow$$
$$\Rightarrow CD \approx 198.855 \, \mathrm{m}.$$
Now using the Law of cosines in $\triangle ACD$, let's calculate $\cos \alpha$:
$$ 200^2=10^2+CD^2-2 \cdot 10 \cdot CD \cdot \cos \alpha \Rightarrow$$
$$\Rightarrow \cos \alpha \approx -0.0896857.$$
Note that $\triangle CEF$ is a right-angled triangle, $CE=r$, and $CF=\frac{CD}{2}$, so:
$$\cos(180^{\circ} - \alpha)= \frac{CF}{CE} \Rightarrow$$
$$\Rightarrow -\cos \alpha =\frac{CD}{2r} \Rightarrow$$
$$\Rightarrow r = \frac{CD}{-2\cos \alpha} \Rightarrow$$
$$\Rightarrow r \approx 1108.6 \, \mathrm{m}.$$
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
Extend $SO$ such that it intersects $C_2$ at $M.$
Now, considering $C_2$, we get $\angle RXY= \angle RMY =\frac{\angle ROY}{2} =\frac{\angle SOY +\angle ROS}{2}.$
On the other hand, considering $C_1$, we have $\angle RXY= \angle SOY$.
Thus, we conclude that $\angle ROS= \angle SOY$. So, $\triangle SOY \cong \triangle ROS.$
We are done.