Question

In the given figure, $∆$ ABC is an equilateral traingle. Points F,D and E are midpoints of side AB, side BC, side AC respectively. Show that $∆$FED is an equilateral traingle.

Answer 1

$∆$ABC is an equilateral triangle.
So, AB = BC = CA
$$\begin{gathered} \Rightarrow \frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{BC}=\frac{1}{2}\mathrm{CA} \\ \Rightarrow \mathrm{FB}=\mathrm{BD}=\mathrm{EC}.....\left(1\right) \end{gathered}$$
Points F,D and E are midpoints of side AB, side BC, side AC respectively.
By midpoint theorem,
$$\begin{gathered} \mathrm{FE}=\frac{1}{2}\mathrm{BC}=\mathrm{BD} \\ \mathrm{DE}=\frac{1}{2}\mathrm{AB}=\mathrm{FB} \\ \mathrm{FD}=\frac{1}{2}\mathrm{AC}=\mathrm{EC} \end{gathered}$$
From (1) we have FB = BD = EC
So, FE = DE = FD
Thus, $∆$FED is an equilateral traingle.