question_answer

  Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. OR Prove that in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.

Answer:

  Given : Two ABC and DBF such that To prove: Construction: Draw  and Proof:                      ?(1) Again, in and , we have                     (Each = 90o)                  (By A.A. rule)                                                 ?(2) [ corresponding sides of similar triangles are proportional] Further,  (Given)                                          ?(3) [ corresponding sides of similar triangles are proportional] From (2) and (3), Putting in (1), we get                                             of (2)] Hence Q.E.D OR Given : A triangle ABC such that AC2 = AB2 + BC2 To prove : Construction: Draw a such that DE = AB, EF = BC and   is right angled at E                                                                                                   [By construction]  AB2 + BC2 = DF2                                                          (By Pythagoras Theorem) But                                    ?(1)       [ DE = AB, EF = BC]                                          (Given) ?(2) Now, in  and          and Hence i.e., Q.E.D