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: Altitude $BE$ and $CF$ are equal.

In $\Delta BEC$ and $\Delta CEB$

$\angle E=\angle F...[Each90°]$

$BC=BC...\left[Common\right]$

$BF=CF...\left[Given\right]$

Therefore, $\Delta BEC\cong \Delta CEB[\mathrm{By}\mathrm{R}.\mathrm{H}.\mathrm{S}]$

So, $\angle C=\angle B...[C.P.C.T]$

Now, $In\Delta ABC,$ $\angle C=\angle B$

Therefore, the triangle$ABC$ is isosceles.