Math, asked by sanchitaniljikar, 6 months ago

BE nad CF are two equal altitudes of triangle ABC ( using RHS rule) prove that triangle ABC is isosceles​

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Answered by ridhima3242
3

Answer:

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Answered by soniabikash
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.

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