Question

BE and CF, the altitudes of $△$ABC are equal. Prove that $△$ABC is an equilateral triangle.

Answer 1

In right triangles CEB and BFC, we have
$\angle BEC=\angle CFB$ (each ${90}^{\circ }$)
BC = BC (common)
BE = CF [Given]
So, by RHS criterion of congruence,
$△BCE\cong △CBF.$

$\Rightarrow \angle B=\angle C[\because \text{corresponding parts of congruent triangles are equal]}$
$\Rightarrow AC=AB....\left(i\right)$
$[\because \text{Sides opposite to equal angles are equal]}$
$\text{Similarly,}△ABD\cong △BAE$
$\Rightarrow \angle B=\angle A$
[Corresponding parts of congruent triangles are equal]
$\Rightarrow AC=BC....\left(ii\right)$
[Sides opposite to equal angles are equal] From (i) and (ii), we get AB = BC = AC
$\text{Hence,}△\angle ABC\text{is an equilateral triangle.}$