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