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:
$△ABC\sim △DEF$
O is a median of BC and P is a median of EF
To Prove:
$\frac{A\left(△ABC\right)}{A\left(△DEF\right)}$ $=\frac{(AO{)}^2}{(DP{)}^2}$
Proof:
Since, $△ABC\sim △DEF$
$\therefore$ $\angle A=\angle D$, $\angle B=\angle E$, $\angle C=\angle F$ (Corresponding Angles of Similar Triangles) $....\left(1\right)$
Also,
$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$ (Corresponding Sides of Similar Triangles) $......\left(2\right)$
Since, $BC=2BO$ and $EF=2EP$
$\therefore$ Equation (2) can be written as,
$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}=\frac{BO}{EP}$ $......\left(3\right)$
In $△AOB$ and $△DPE$
$\angle B=\angle E$ (From 1)
$\frac{AB}{DE}=\frac{BO}{EP}$ (From 3)
$\therefore$ By SAS Criterion of Similarity, $△AOB$ $\sim$$△DPE$
$\therefore$ $\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}=\frac{AO}{DP}=$Ratio of their heights $....\left(4\right)$ (Corresponding Sides of Similar Triangles)
$\frac{A\left(△ABC\right)}{A\left(△DEF\right)}$ $=\frac{\frac{1}{2}\times BC\times Height}{\frac{1}{2}\times EF\times Height}$$=\frac{(AO{)}^2}{(DP{)}^2}$