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

In $\Delta BCF$ and$\Delta CBE$,

$\angle BFC=\angle CEB$ (Each $90º$)

Hyp. $BC=$ Hyp. $BC$ (Common Side)

Side $FC=$ Side $EB$ (Given)

$\therefore$ By R.H.S. criterion of congruence, we have

$\Delta BCF\cong \Delta CBE$

$\therefore \angle FBC=\angle ECB$ (CPCT)

In $\Delta ABC$,

$\angle ABC=\angle ACB$

[$\because \angle FBC=\angle ECB$]

$\therefore AB=AC$ (Converse of isosceles triangle theorem)

$\therefore$ $\Delta ABC$ is an isosceles triangle.