Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of th

Question 6 Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution:

Given:

AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF

We have to prove: $\dfrac{Area(ΔABC)}{Area(ΔDEF)} = \dfrac{AM^2}{DN^2}$

Since, ΔABC ~ ΔDEF (Given)

$\dfrac{ Area(ΔABC)}{Area(ΔDEF)} = \dfrac{AB^2}{DE^2}$ ……………………………(i)

and,$\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{CA}{FD} $………………………………………(ii)

$\dfrac{AB}{DE} = \dfrac{\frac{1}{2}\times BC}{\frac{1}{2}\times EF} = \dfrac{CD}{FD}$

In ΔABM and ΔDEN,

Since ΔABC ~ ΔDEF

$\angle B = \angle E$

$\dfrac{AB}{DE} = \dfrac{BM}{EN}$ [Already Proved in equation (i)]

ΔABC ~ ΔDEF [SAS similarity criterion]

$\dfrac{AB}{DE} = \dfrac{AM}{DN}$ …………………………………………………..(iii)

ΔABM ~ ΔDEN

As the areas of two similar triangles are proportional to the squares of the corresponding sides

$\dfrac{ area(ΔABC)}{area(ΔDEF)} = \dfrac{AB^2}{DE^2} = \dfrac{AM^2}{DN^2}$

Hence, proved

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