Question

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

Answer 1

In right triangles BCE and BFC, we have
$\angle BEF=\angle BFC$ $[each{90}^{\circ }$]
Hyp. BC = Hyp. BC
BE = CF [Given]
So, by RHS criterion of congruence,

$\Delta BCE\cong \Delta CBF.$
$\Rightarrow \angle B=\angle C$
[Corresponding parts of Congruent triangles are equal]
$\Rightarrow AC=AB$ ....(i)
[Sides opposite to equal angles are equal]
Similarly, $\Delta ABD\cong \Delta BAE$
$\Rightarrow \angle B=\angle A$
[Corresponding parts of congruent triangles are equal]
$\Rightarrow AC=BC$ ....(ii)
[Sides opposite to equal angles are equal]
From (i) and (ii), we get
AB = BC = AC
Hence, $\Delta ABC$ is an equilateral triangle.