Question

BEand CF are two equal altitudes of a triangle ABC.using rhs congruence rule, prove that the triangle ABC is isosceles triangle

Answer 1

Prove by taking triangle ABF and triangle ACF

Angle a = angle a (common)

Angle AEF = Angle BFA

By AAA

We can tell they are congruent

By cpct ab = ac

As opp sides are equal they are 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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