A) \[AE\]
B) \[CF\]
C) \[BE\]
D) \[CE\]
Correct Answer: D
Solution :
In \[\Delta ABC,AB=AC\](given) E and F are respectively the mid-points of the sides AB and AC. \[\Rightarrow \]\[AE=\frac{1}{2}AB\]and \[AF=\frac{1}{2}AC\] We know that halves of the equal sides are equal \[\Rightarrow \]\[AE=AF\] ?..(i) Now, in \[\Delta ABF\]and \[\Delta ACF,\] \[AB=AC\](Given) \[\angle BAF=\angle CAE\] (Each\[=\angle A.\]) \[AE=AF\] [By (i)] Thus, \[\Delta ABF\cong \Delta ACE\](By S.A.S. congruence) \[\therefore \]\[BF=CE(c.p.c.t)\]
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