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

Explanation

Given : `triangleABC~triangleDEF` AP and DQ are the mdians of `triangleABC and triangleDEF` respectively.
To Prove: `(ar(triangleABC))/(ar(triangleDEF))= (AP^(2))/(DQ^(2))`
Proof : AP and DQ are medians
Therefore, BP=PC and EQ=QF
and given ` triangleABC ~ triangleDEF`
Therefore, `(AB)/(DE) = (AC)/(DF)= (BC)/(EF) and angleA = angleD, angleB=angleE and angleC= angleF`
Now `(AB)/(DE)= (BC)/(EF)`
` Rightarrow (AB)/(DE)= (2BP)/(2EQ) Rightarrow (AB)/(DE)= (BP)/(EQ)` ...(1)
and ` angleB= angleE`
Hence ` triangleABP ~ triangleDEQ` (by SAS similarity)
Now in `triangleABC and triangleDEF`
`(ar(triangleABC))/(ar(triangleDEF)) = (AB^(2))/(DE^(2))`
( the ratio of areas of two similar `triangle's` is equal to ratio of squares of their corresponding sides)
form (1) and (2) , we have
`(ar(triangleABC))/(ar(triangleDEF))= (AP^(2))/(DQ^(2))` Hence proved