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According to the question :
ΔABC = an isosceles triangle
AB = AC
=> ΔACB = ΔABC
[ Angles are opposite to equal sides of a triangle Δ are equal ]
=> ΔBCE = ΔCBF
To find :
In ΔBEC and ΔCFB, we have
=> ΔBCE = ΔCBF [ Proved ]
=> ΔBEC = ΔCFB [ Each 90° ]
=> ΔBC = CB [ common ]
=> ΔBEC ≈ ΔCFB [By ASA Congruent axiom ]
=> BE = CF [ By corresponding part of congruent triangles ]
Hence, Proved
♥ So It's Done ♥ !!
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