# The areas of two similar triangles are in the ratio of the square of the corresponding medians.

## Solution :

Given : $$\triangle$$ ABC ~ $$\triangle$$ DEF and AP, DQ are their medians.

To Prove : $$area(\triangle ABC)\over area(\triangle DEF)$$ = $${AP}^2\over {DQ}^2$$

Proof : Since the ratio of the area of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

$$\therefore$$  $$area (\triangle ABC)\over area (\triangle DEF)$$ = $${AB}^2\over {DE}^2$$            ………..(1)

Now,  $$\triangle$$ ABC ~ $$\triangle$$ DEF

$$\implies$$  $$AB\over DE$$ = $$BC\over EF$$ = $$2BP\over 2EQ$$ = $$BP\over EQ$$               ………….(2)

Thus, $$AB\over DE$$ = $$BP\over EQ$$ and $$\angle$$ B = $$\angle$$ E             (because $$\triangle$$ ABC ~ $$\triangle$$ DEF)

By SAS similarity,

$$\triangle$$ APB ~ $$\triangle$$ DQE

So, $$BP\over EQ$$ = $$AP\over DQ$$             ……….(3)

From (2) and (3), we get

$$AB\over DE$$ = $$AP\over DQ$$     $$\implies$$  $${AB}^2\over {DE}^2$$ = $${AP}^2\over {DQ}^2$$            ………(4)

Substituting the value of $${AB}^2\over {DE}^2$$ from (4) in (1), we get

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