Question

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

Answer 1

ur answer is in attachment

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Answer 2

$if \: you \: see \: the \: attachment \: \\ you \: can \: understand \: clearly \: brother$

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.