$ABC$ 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.

Given:

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

To do:

We have to prove that the triangle is isosceles.

Solution:

In $\triangle ABC, D$ is mid-point of $BC$

$DE \perp AB, DF \perp AC$ and $DE = DF$

In $\triangle BDE$ and $\triangle CDF$,

$DE = DF$

$BD = CD$

Therefore, by RHS axiom,

$\triangle BDE \cong \triangle CDF$

This implies,

$\angle B = \angle C$

In $\triangle ABC$,

$\angle B = \angle C$

This implies,

$AC = AB$                 (Sides opposite to equal angles are equal)

Therefore,

$\triangle ABC$ is an isosceles triangle.

Hence proved.

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Updated on: 10-Oct-2022

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