In a triangle ABC median AM=17,altitude AD=15 and the circum radius R=10find BC square

E is the perpendicular on AD from O. We can show by side-side-side criteria that triangles OMBand OMC are congruent, hence angle OMC =90. So OE=MD={17^2-15^2}^1/2=8, andOM=ED=AD-AE=15-\sqrt{10^2-8^2}=9. Thus, BC^2=4BM^2=4(10^2-9^2)=76.

Regards

Sumit