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

Explanation

Given : `DeltaABC` in which BE and CF are the altitudes such that `BE=CF`. To prove : `DeltaABC` is an isosceles triangle i.e., `AB = AC`.
Proof : Since BE and CF are the altitudes
`:. angle1 = angle2" "("each "90^(@))`
Now, in `DeltaBFC` and `DeltaBEC`,
`:' {(angle1=angle2,("given")),(BC=BC,("common")),(CF=BE,("given")):}`
`:. DeltaBFC cong DeltaBEC` (RIIS)
`:. angleB=angleC` (c.p.c.t)
`implies AC = AB` (sides opposite to equal angles are equal)
`:. DeltaABC` is an isosceles triangle. Hence Proved.
We can also prove this by taking AAS congruence rule in `Deltas` ABE and AFC, if no particular method is asked)