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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 c2 = a2 + b2 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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Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. |