Question

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

Answer 1

Given that the altitudes of a triangle $ABC$ are equal

i.e. $AD=BE=CF$

Area of $\Delta ABC=\frac{1}{2}\left(BC\right)\times \left(AD\right)$

Area of $\Delta ABC=\frac{1}{2}\left(AB\right)\times \left(CF\right)$

Area of $\Delta ABC=\frac{1}{2}\left(AC\right)\times \left(BE\right)$

$\therefore \frac{1}{2}\left(BC\right)\times \left(AD\right)=\frac{1}{2}\left(AB\right)\times \left(CF\right)=\frac{1}{2}\left(AC\right)\times \left(BE\right)$

But $AD=BE=CF$ then

$\Rightarrow \left(BC\right)\times \left(AD\right)=\left(AB\right)\times \left(AD\right)=\left(AC\right)\times \left(AD\right)$

$\Rightarrow BC=AB=AC$

So three sides of the triangle are same.

Thus, the triangle $ABC$ is an equilateral triangle.