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

Hint: -First you have to draw a diagram so that you can understand what has to be proved. Use the property of a similar triangle that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Complete step-by-step solution -
Given: -
Let $\mathrm{\Delta }\text{ABC}\sim \mathrm{\Delta }\text{DEF}$
Here AM is median
Hence BM=CM=${\displaystyle \frac{1}{2}}\text{BC}$
Similarly, DN is median
Hence EN=FN=${\displaystyle \frac{1}{2}}\text{EF}$
To prove: ${\displaystyle \frac{ar\left(\mathrm{\Delta }\text{ABC}\right)}{ar\left(\mathrm{\Delta }\text{DEF}\right)}}={\displaystyle \frac{\text{A}{\text{M}}^2}{\text{D}{\text{N}}^2}}$
Proof:$\mathrm{\Delta }\text{ABC}\sim \mathrm{\Delta }\text{DEF}$$\left(given\right)$
The property of similar triangle ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
$\therefore {\displaystyle \frac{ar\left(\mathrm{\Delta }\text{ABC}\right)}{ar\left(\mathrm{\Delta }\text{DEF}\right)}}={\displaystyle \frac{\text{A}{\text{B}}^2}{\text{D}{\text{E}}^2}}\dots \left(1\right)$
It is also a property of similar triangles that corresponding sides of similar triangles are in the same proportion.
So, ${\displaystyle \frac{\text{AB}}{\text{DE}}}={\displaystyle \frac{\text{BC}}{\text{EF}}}={\displaystyle \frac{\text{CA}}{\text{FD}}}\dots \left(2\right)$
$\Rightarrow {\displaystyle \frac{\text{AB}}{\text{DE}}}={\displaystyle \frac{{\displaystyle \frac{1}{2}}\text{BC}}{{\displaystyle \frac{1}{2}}\text{EF}}}={\displaystyle \frac{\text{BM}}{\text{EN}}}$$\dots \dots \left(3\right)$
In $\mathrm{\Delta }\text{ABM}$and $\mathrm{\Delta }\text{DEN}$, we have
$\mathrm{\angle }B=\mathrm{\angle }E\left[\because \mathrm{\Delta }\text{ABC}\sim \mathrm{\Delta }\text{DEF}\right]$(corresponding angle of similar triangle are equal)
${\displaystyle \frac{\text{AB}}{\text{DE}}}={\displaystyle \frac{\text{BM}}{\text{EN}}}\left[\text{proved in }\left(3\right)\right]$
$\therefore \mathrm{\Delta }\text{ABM}\sim \mathrm{\Delta }\text{DEN by SAS similarity criterian}\text{.}$
$\therefore {\displaystyle \frac{\text{AB}}{\text{DE}}}={\displaystyle \frac{\text{AM}}{\text{DN}}}$$\dots \dots \left(4\right)$ (corresponding sides of similar triangle are proportional)
We know the ratio areas of two similar triangles are equal to the squares of the corresponding sides.
$\therefore {\displaystyle \frac{ar\left(\mathrm{\Delta }\text{ABC}\right)}{ar\left(\mathrm{\Delta }\text{DEF}\right)}}={\displaystyle \frac{\text{A}{\text{B}}^2}{\text{D}{\text{E}}^2}}={\displaystyle \frac{\text{A}{\text{M}}^2}{\text{D}{\text{N}}^2}}$ (using equation 4 $\left({\displaystyle \frac{\text{AB}}{\text{DE}}}={\displaystyle \frac{\text{AM}}{\text{DN}}}\right)$)
Hence proved.

Note: -The key concept of solving is first draw a diagram and write what is given and what has to be proved. Then use the properties of a similar triangle to prove the question. You should have remembered all the properties.