Question

In ∆ABC, D and E are the midpoints of AB and AC, respectively. Find the ratio of the areas of ∆ADE and ∆ABC.

Answer 1

It is given that D and E are midpoints of AB and AC.
Applying midpoint theorem, we can conclude that DE $\parallel$ BC.
Hence, by B.P.T., we get:
$\frac{AD}{AB}=\frac{AE}{AC}$
Also, $\angle A=\angle A$.

Applying SAS similarity theorem, we can conclude that $△ADE~△ABC$.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
$$\begin{gathered} \therefore \frac{\mathrm{ar}(∆ADE)}{\mathrm{ar}(∆ABC)}=\frac{D{E}^2}{B{C}^2} \\ =\frac{{\left({\displaystyle \frac{1}{2}}BC\right)}^2}{B{C}^2} \\ =\frac{1}{4} \end{gathered}$$