Question

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

Answer 1

Given $BE$ and $CF$ are two equal altitudes of triangle $ABC$

Given $BE$ is a altitude

$\therefore \angle BEC={90}^0$

And $CF$ is a altitude

$\therefore \angle BFC={90}^0$

In $\Delta BEC$ and $BFC$

$BE=CF$ ...Given

$BC=BC$ ...Common

$\angle BFC=\angle BEC={90}^0$ ...(Proved above)

$\therefore \Delta BEC\cong BFC$ ...RHS test of Congruence

$\therefore BE=CF$

The $E$ and $F$ are mid points of $AB$ and $AC$

$\therefore AB=AC$

Then, triangle $ABC$ is an isosceles triangle.