Question

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

Answer 1

here is the solution====

Consider two triangles ABC and DEF.

AX and DY are the bisectors of the angles A and D respectively.



Ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. so,

Area (ΔABC) / Area (ΔDEF) = AB2/ DE2 -----(1)

ΔABC ~ ΔDEF ⇒ ∠A = ∠D

1/ 2 ∠A = 1 / 2 ∠D ⇒ ∠BAX = ∠EDY

Consider ΔABX and ΔEDY

∠BAX = ∠EDY

∠B = ∠E

So, ΔABX ~ ΔEDY [By A-A Similarity]

AB/DE = AX/DY

⇒ AB2/DE2 = AX2/DY2 --------- (2)

From equations (1) and (2), we get

Area (ΔABC) / Area (ΔDEF) = AX2/ DY2

Hence proved.

Please mark as brainliest....