Question

D,E F are the midpoints of the sides AB,BC CA of an isoscels triangle ABC in which AB=BC. prove that the triangle DEF is also an isoscels triangle.

Answer 1

$$\begin{gathered} △ABC\hspace{0.17em}isisosceleswithAB=BC \\ \Rightarrow AD=BD=BE=CE \end{gathered}$$

Also,
$$\begin{gathered} \because DandEaremidpointsofABandBCrespectively, \\ DE\parallel AC \end{gathered}$$
$$\begin{gathered} \Rightarrow \angle BDE=\angle A,and,\angle BED=\angle C \\ \Rightarrow \angle BDE=\angle BED\left[\because \angle A=\angle CasAB=BC\right]------\left(1\right) \end{gathered}$$

Similarly,
$$\begin{gathered} \because EandFaremidpointsofBCandACrespectively, \\ FE\parallel AB \end{gathered}$$
$\Rightarrow \angle FEC==\angle B------\left(2\right)$

Again,
$$\begin{gathered} \because DandFaremidpointsofABandACrespectively, \\ DF\parallel BC \end{gathered}$$
$\Rightarrow \angle ADF=\angle B-------\left(3\right)$

From equations (2) and (3),
$\angle FEC=\angle ADF------\left(4\right)$

Eq (1) + Eq (4)$$\begin{gathered} \Rightarrow \angle BDE+\angle ADF=\angle BED+\angle FEC \\ \Rightarrow 180°-\angle FDE=180°-\angle FED \\ \Rightarrow \angle FDE=\angle FED \\ \Rightarrow △DEFisisosceleswithDF=EF \end{gathered}$$

Hence Proved.