Question

I triangle ABC altitudes be and cf are equal . Prove that triangle ABC is an isosceles triangle.

Answer 1

in ∆ABE and ∆ACE,
angle A=angle A (common)
BE =CF (given)
angle AFC=angle AEB (90°)
therefore AB =AC(by CPCT)
thus ABC is an isosceles triangle

Answer 2

Hello mate ^_^

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Solution:

In ∆BEC and ∆CFB

BE=CF                (Given)

∠BEC=∠CFB              (Each given equal to 90°)

BC=CB                (Common)

Therefore, by RHS rule, ∆BEC≅∆CFB

It means that ∠C=∠B        (Corresponding parts of congruent triangles are equal)

⇒AB=AC                (In a triangle, sides opposite to equal angles are equal)

Therefore, ∆ABC is isosceles.

hope, this will help you.

Thank you______❤

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