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.triangles

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

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

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