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

In \(\triangle{BEC}\) and \(\triangle{CBF}\), we have,
\(\angle{BEC}\) = \(\angle{CFB}\) = \(90^\circ\) ...(Given)
BE = CF ...(Given)
and BC is the Common side.
\(\therefore \) \(\triangle{BEC}\) \(\displaystyle \cong \) \(\triangle{CFB}\) ...(By RHS congruency test)
Thus, EC = FB ...(i)(By CPCT)

In \(\triangle{AEB}\) and \(\triangle{AFC}\), we have,
\(\angle{AEB}\) = \(\angle{AFC}\) = \(90^\circ\) ...(Given)
\(\angle{A}\) is Common angle.
and EB = FC ...(Given)
\(\therefore \) \(\triangle{AEB}\) \(\displaystyle \cong \) \(\triangle{AFC}\) ...(By AAS congruency test)
Thus, AE = AF ...(ii)(By CPCT)

Now, on adding Equation (i) and (ii),
we get,

EC + AE = FB + AF
AC = AB ...(Since, AC = EC + AE and AB = FB + AF)

Hence, it is proved that, \(\triangle{ABC}\) is an isosceles triangle.