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04/06/2021 11:44 am
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Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
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04/06/2021 11:54 am
AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF.
Area(ΔABC)/Area(ΔDEF) = AM2/DN2
Since, ΔABC ~ ΔDEF (Given)
∴ Area(ΔABC)/Area(ΔDEF) = (AB2/DE2) ……………………(i)
and, AB/DE = BC/EF = CA/FD …………………(ii)
\(\frac{AB}{DE} = \frac{\frac{1}{2}BC}{\frac{1}{2}EF} = \frac{CD}{FD}\)
In ΔABM and ΔDEN,
Since ΔABC ~ ΔDEF
∴ ∠B = ∠E
AB/DE = BM/EN [Already Proved in equation (i)]
∴ ΔABC ~ ΔDEF [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
Hence, proved.
This post was modified 4 years ago by Raavi Tiwari