question_answer

  In a right angled triangle, the square of hypotenuse is equal to sum of squares on other two sides, prove. In the given figure, perpendiculars OD, OE and OF are drawn to sides BC, CA and AB respectively from a point 0 in the interior of a . Prove that : AF2 + BD2 + CE2 == AE2 + CD2 + BF2.

Answer:

  From right triangles and we get and                    ...(1) Similarly, we have                                ?(2) and                       ?(3) Adding (1), (2) and (3), we get                                                  (OB2 - OC2) + (OC2 - OA2) + (OA2 - OB2) = (BD2 - CD2) + (CE2 - AE2) + (AF2 - BF2) 0 = (BD2 + CE2 + AF2) - (AE2 + CD2 + BF2) AF2 + BD2 + CE2 = AE2 + BF2 + CD2