Question

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

Answer 1

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

Answer 2

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