



 
Amit, I folded the sheet of paper CDBP along the diagonal CB so that P moved to A. then drew a line segment EF to meet BC in a right angle at F. Let CA = a, AB = b and then BC = c where c^{2} = a^{2} + b^{2} by Pythagoras' Theorem.
By the symmetry F is the midpoint of BC. Let EF = h, the height of triangle CEB. Triangles ABC and EBF are similar. Can you complete the problem now?  


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