Question is as per title.
Here is a sketch that I made :
By Pythagorean theorem , DM is 8. Now how can I calculate BD and MC? I still haven't found a way to utilize the information that the circum-radius is 10. (I have drawn circum-circle inside but it may be outside)
Any hints are apreciated. (This is not class-homework , I'm solving sample questions for a competitive exam )
This solution is due to r9m's insight given on the chat. But since he does not want to give an answer, I am giving his wonderful geometric solution:
$E$ is the perpendicular on $AD$ from $O$. We can show by side-side-side criteria that triangles $OMB$ and $OMC$ are congruent, hence $\angle OMC =90^{\circ}$. So $OE=MD=\sqrt{17^2-15^2}=8$, and $OM=ED=AD-AE=15-\sqrt{10^2-8^2}=9$. Thus, $BC^2=4BM^2=4(10^2-9^2)=76$.
My original ugly solution:
Note that: $$2A=\frac{abc}{2R}=ah$$
Substituing the values gives, $bc=300$. Since, $2A=bc\cdot\sin A$, so $2bc\cos A=600\sqrt{1-(\frac{A}{150})^2}=600\sqrt{1-\frac{a^2}{400}}$.
Now by Apollonius Theroem, $578+\frac{a^2}2=b^2+c^2$. Thus by cosine rule, $600\sqrt{1-\frac{a^2}{400}}=578-\frac{a^2}2$. Squaring will give us a quadratic in $a^2$, which you can solve to get the answer.