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

The question is to prove the ratio of the areas of two similar triangles is equal to the square of their corresponding sides not square of ratio of corresponding sides.

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how are you guys area theorms so as i cant type much you can refer to text books i can give you some tips to win in exam first when teacher is away exchange your paper with someone sitting near you without seeing write the answer on your table with pencil so that you can refer while writing exam

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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.

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Given: AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF.
To Prove: area(ΔABC)/area(ΔDEF) = AM2/DN2
Proof: ΔABC ~ ΔDEF (Given)
∴ area(ΔABC)/area(ΔDEF) = (AB2/DE2) ...(i)
and, AB/DE = BC/EF = CA/FD ...(ii)
In ΔABM and ΔDEN, we have
∠B = ∠E [Since ΔABC ~ ΔDEF]
AB/DE = BM/EN [Prove in (i)]
∴ ΔABC ~ ΔDEF [By SAS similarity criterion]
⇒ AB/DE = AM/DN ...(iii)
∴ ΔABM ~ ΔDEN
As the areas of two similar triangles are proportional to the squares of the corresponding sides.
∴ area(ΔABC)/area(ΔDEF) = AB2/DE2 = AM2/DN2

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