If two triangles are similar show …

Let the two similar triangles ABC and DEF in which AP and DQ are the medians respectively.

we have to prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

If ΔABC~ΔDEF

{tex}⇒ \frac{ar(BCA)}{ar(EFD)}=\frac{AB^2}{DE^2} {/tex} →   (1)

As ΔABC~ΔDEF

{tex}\frac{AB}{DE}=\frac{BC}{EF}=\frac{2BP}{2EQ}{/tex}

Hence,{tex} \frac{AB}{DE}=\frac{BP}{EQ}{/tex}

In ΔABP and ΔDEQ

{tex}\frac{AB}{DE}=\frac{BP}{EQ}{/tex}

∠B=∠E   (∵ΔABC~ΔDEF)

By SAS rule, ΔABP~ΔDEQ

{tex}⇒ \frac{AB}{DE}=\frac{AP}{DQ}{/tex}

Squaring, we get

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

Comparing (1) and (2), we get

{tex}\frac{ar(BCA)}{ar(EFD)}=\frac{AP^2}{DQ^2} {/tex} 

Hence, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.