A) CF
B) AB
C) CE
D) BF
Correct Answer: A
Solution :
Since, AB = AC \[\Rightarrow \] \[\frac{1}{2}AB=\frac{1}{2}AC\] \[\Rightarrow \] \[BF=EC\] Also. \[AB=AC\text{ }\Rightarrow \text{ }\angle B=\angle C\] [Angles opposite to equal sides are equal] In \[\Delta \text{ }BEC\]and \[\Delta \text{ }CFB\] EC = FB (proved above) \[\angle B=\angle C~\] (proved above) BC = BC (common) \[\therefore \] \[\Delta \,BEC\cong \Delta \,CFB\] (By SAS) \[\Rightarrow \] \[BE=CF\] (By CPCT)You need to login to perform this action.
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