Question

BE and CF are two equal altitudes of a triangle ABC. Using RHS

congruence rule, prove that the triangle ABC is an isosceles triangle.​

Answer 1

Answer:

In ΔBCF andΔCBE,

∠BFC=∠CEB (Each 90º)

Hyp. BC= Hyp. BC (Common Side)

Side FC= Side EB (Given)

∴ By R.H.S. criterion of congruence, we have

ΔBCF≅ΔCBE

∴∠FBC=∠ECB (CPCT)

In ΔABC,

∠ABC=∠ACB

[∵∠FBC=∠ECB]

∴AB=AC (Converse of isosceles triangle theorem)

∴ ΔABC is an isosceles triangle.

( I hope it will help you, Thankyou )

Answer 2

To Find:-

To prove that ∆ABC is an Isosceles Triangle by using R.H.S congruency rule.

Given:-

BE and CF are two equal altitudes of a ∆ABC

Solution:-

step-by-step explanation:

BE is an altitude

Then, ∠ AEB = ∠ CEB = 90°

Also,

CF is an altitude

∠AFC = ∠BFC = 90°

BE = CF

Now,

In ∆BCF and ∆CBE,

∠ BFC and ∠ CBE

BC = CB ( common )

FC = EB

∆BCF is congruent to ∆CBE ( R.H.S. congruency rule )

∠FBC = ∠ECB ( C.P.C.T. )

∠ABC = ∠ACB

AB = AC ( sides opposite to equal sides are equal )

Hence, Proved

So, ∆ABC is an Isosceles Triangle.