$ABC$ is a triangle in which $BE$ and $CF$ are respectively, the perpendiculars to the sides $AC$ and $AB$. If $BE = CF$, prove that $\triangle ABC$ is isosceles.

Given:

$ABC$ is a triangle in which $BE$ and $CF$ are respectively, the perpendiculars to the sides $AC$ and $AB$ and $BE = CF$.

To do:

We have to prove that $\triangle ABC$ is isosceles.

Solution:

In $\triangle ABC$,

$BE \perp AC$ and $CF \perp AB$

$BE = CF$

In $\triangle BCE$ and $\triangle BCF$

$BE = CF$

$BC = BC$        (Common)

Therefore, by RHS axiom,

$\triangle BCE \cong \triangle BCF$

This implies,

$\angle BCE = \angle CBF$                   (CPCT)

$AB = AC$                   (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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