BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:

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In $\triangle B E C$ and $\triangle C F B$,

$\angle B E C=\angle C F B\left(\right.$ Each $\left.90^{\circ}\right)$

$\mathrm{BC}=\mathrm{CB}$ (Common)

$\mathrm{BE}=\mathrm{CF}$ (Given)

$\therefore \Delta \mathrm{BE} \mathrm{C} \cong \triangle \mathrm{CFB}(\mathrm{By} \mathrm{RHS}$ congruency $)$

$\Rightarrow \angle B C E=\angle C B F($ By CPCT $)$

$\therefore A B=A C$ (Sides opposite to equal angles of a triangle are equal)

Hence, $\triangle \mathrm{ABC}$ is isosceles.
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