Question

BE and CF are two equal altitudes of a triangle ABC. Using the RHS congruence rule, prove that the triangle ABC is isosceles​
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Answer 1

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. hope it helps you please mark me as brainliest please by army

Answer 2

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.