Krishna
0

Step 1: Make a suitable diagram for this question. Note down the main points

              


                            GIVEN: \triangle ABC and \triangle DEF are similar triangles  

                    

                           We have to prove   \frac{area \triangle ABC}{ area \triangle DEF} = \frac{(AP)^2}{(DQ)^2}


Step 2: Find the ratio of the areas of two similar triangles

              NOTE: \frac{area \triangle ABC}{ area \triangle DEF} = \frac{(AB)^2}{(DE)^2}............................................(1)

              ( Theorem: The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides)

        

                      \frac{AB}{DE} = \frac{BC}{EF} ( \because \triangle ABC\ and\ \triangle DEF are similar triangles)

                     \frac{AB}{DE} = \frac{2*BP}{2*EQ} ( \because  P, Q are mid points)

                     \frac{AB}{DE} = \frac{BP}{EQ}...........................(2)


Step 3:  Prove that the triangles  ABP and DEQ are similar.

        EXPLANATION:   \frac{AB}{DE} = \frac{BP}{EQ} ( \becauseequation (2))

                                       \angle B = \angle Q   ( \because△ ABC ~ △ DEF )


                So, we can write   \frac{AB}{DE} = \frac{AP}{DQ}   

                                           Square on both sides

                                              \frac{(AB)^2}{(DE)^2} = \frac{(AP)^2}{(DQ)^2} .....................................(3)


Step 4:  Compare equation (1) and (2)

                   \frac{area \triangle ABC}{ area \triangle DEF} = \frac{(AB)^2}{(DE)^2}

                       \frac{(AB)^2}{(DE)^2} = \frac{(AP)^2}{(DQ)^2}

                From equation (1) and (2) we can write

                   \frac{area \triangle ABC}{ area \triangle DEF} = \frac{(AP)^2}{(DQ)^2}

                  Hence proved