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