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

Given In ∆ ABC, BE = CF and ∠BFC =∠CEB = 90°

To prove ∆ ABC is an isosceles triangle.

Proof In ∆BEC and ∆CFB

BE =CF

∠ BEC = ∠ CFB

and BC = BC

⇒ ∠ ECB =∠ FBC '[byCPCT]

or ∠ ACB =∠ ABC

⇒ AB = AC

∆ ABC is an isosceles triangle.