Question

ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is isosceles.

Answer 1

Given $\Delta ABC,\text{D is mid-point of BC and DE}\perp AB,DF\perp ACandDE=DF$

To Prove : $\Delta ABC$ is an isosceles triangle

Proof : In right $\Delta BDEand\Delta CDF,$

Side DE = DF

Hyp. BD = CD

$\therefore \Delta BDE\cong \Delta CDF\text{(RHS axiom)}$

$\therefore \angle B=\angle C(c.p.c.t.)$

Now in $\Delta ABC,$

$\angle B=\angle C\left(Prove\right)$

$\therefore AC=AB\text{(Sides opposite to equal angles)}$

$\therefore \Delta ABC\text{is an isosceles triangle}$