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

Asked by Pragya Singh | 1 year ago |  66

##### 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)

In ΔABM and ΔDEN,

Since ΔABC ~ ΔDEF

∠B = ∠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.

Answered by Abhisek | 1 year ago

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