$ \mathrm{BE} $ and $ \mathrm{CF} $ are two equal altitudes of a triangle $ \mathrm{ABC} $. Using RHS congruence rule, prove that the triangle $ \mathrm{ABC} $ is isosceles.

Given:

$BE$ and $CF$ are two equal altitudes of a triangle $ABC$.

To do:

By using the RHS congruence rule we have to prove that the triangle  $ABC$ is isosceles.

Solution:

Let us consider $\triangle BEC$ and $\triangle CFB$,

We have,

$BE$ and $CF$ are two equal altitudes of a triangle $ABC$.

This implies,

$\angle BEC=\angle CFB=90^o$

and $BE=CF$

Since $BC$ is the common side

we get,

$BC=CB$

We know that According to the RHS rule if the hypotenuse and one side of a right-angled triangle are equal to the corresponding hypotenuse and one side of another right-angled triangle; then both the right-angled triangle are said to be congruent.
Therefore,

$\triangle BEC \cong \triangle CFB$

We also know that,

The corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding angles and sides must be equal.

This implies,

$\angle C=\angle B$

Therefore,

As the sides opposite to equal angles is always equal we get,

$AB=AC$.

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Updated on: 10-Oct-2022

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