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

In ΔFBC and ΔECB,
∠BFC = ∠CEB [∵ Each 90°]
BC = BC [∵ Common]
FC = BE [∵ Given]
Hence, ΔFBC ≅ ΔECB [∵ RHS Congruency Rule]
∠FBC = ∠ECB [∵ CPCT]
⇒ AC = AB [∵ Angles opposite to equal Sides are equal]
Hence, ΔABC. is an isosceles triangle.