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