Question

∆ABC and ∆BDE are two equilateral triangles such that D is the midpoint of BC. Then, prove that $\mathrm{ar}(∆BDE)=\frac{1}{4}\mathrm{ar}(∆ABC)$.

Answer 1

∆ABC and ∆BDE are two equilateral triangles and D is the mid point of BC.​
Let AB = BC = AC = a
Then BD = BE = ED = $\frac{a}{2}$
We know that the area of an equilateral triangle is given by $\frac{\sqrt{3}}{4}(\mathrm{side}{)}^2$.
So, ar(∆ABC ) = $\frac{\sqrt{3}}{4}A{B}^2=\frac{\sqrt{3}}{4}{a}^2$ ...(i)
Also, ar(∆BDE ) = $\frac{\sqrt{3}}{4}\times {\left({\displaystyle BD}\right)}^2=\frac{\sqrt{3}}{4}\times {\left(\frac{a}{2}\right)}^2=\frac{\sqrt{3}}{4}\times \frac{a^2}{4}=\frac{1}{4}\left(\frac{\sqrt{3}}{4}{a}^2\right)$
From (i), we have:
ar(∆BDE ) = ar(∆ABC)
Hence, proved.