Question

BE and CF are two equal altitudes of a triangle ABC. using RHS congruence rule prove that ABC is isosceles​

Answer 1

Answer:

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.

Answer 2

Explanation:

see the first ans in your question