\(BE\) and \(CF\) are two equal altitudes of a triangle \(ABC\). Using \(RHS\) congruence rule, prove that the triangle \(ABC\) is isosceles.
Proof:
In \(ΔBEC\) and \(ΔCEB\),
\(∠E=∠F=90^{\circ}\)
\(BC=BC\) -[common]
\(BF=CF\) [Given]
Therefore by rule, \(ΔBEC ≅ ΔCEB\)
\(∠C=∠B\) [C.P.C.T]
In ΔABC, \(∠C=∠\)
That is, \(ABC\) is an isosceles triangle.
Hence, we proved.