Question

In triangle abc median be and cf are equal prove that abc is an isosceles triangle

Answer 1

Lets take the triangle BCF and triangle EBC
FC = EB
Angle ECB = Angle FBC
BC is the common side
So triangle ECB and triangle FBC are congruent
FB = EC
And AF = AE
SO FB +AF = EC+ AE
AB = AC ( PROVED)

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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