Question

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BE and CF are two equal altitudes of a triangle ABC. Prove that the triangle ABC is an isosceles.​

Answer 1

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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)

∴AB=AC (Converse of isosceles triangle theorem)

∴ ΔABC is an isosceles triangle.

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Answer 2

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• BE and CF are two equal altitudes of a triangle ABC. Prove that the triangle ABC is an isosceles.

$\huge \mathbb\red {answer}$

• 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.
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