Question

Be and CF are two equal altitudes of a triangle ABC using RHS congruence rule prove that the triangle ABC is an isosceles

Answer 1

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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Answer 2

hii friend

here's your answer looking for

Consider triangle BFC AND CEB

BC = CB. ( COMMON )

ANGLE BFC = ANGLE CEB = 90 (GIVEN)

CF = BE. ( GIVEN )

SO, triangle BFC is congruent to triangle CEB. ( RHS )

Hence , BF = CE. ( CPCT )_____________(1)

and, considered triangle ABE and ACF

angle BAC = angle CAB. ( COMMON )

angle AFC = angle AEB. (GIVEN)

CF = BE. (GIVEN)

HENCE,. triangle ABE is congruent to triangle ACF. (AAS)

So, AF = AE. (CPCT). _______________(2)

adding (1) and (2)

we get,

BF + AF = CE + AE

AB = AC

hence , triangle ABC is iso. triangle.

hope it's helpful to you

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From brainly star soniabikash