A) \[AF=\frac{3}{2}AB\]
B) \[AF=2AB\]
C) \[AF=3AB\]
D) \[A{{F}^{2}}=2A{{B}^{2}}\]
Correct Answer: B
Solution :
(b): In \[\Delta EDC\] and \[\Delta EFB\], we have: \[\angle DCE=\angle EBF\] \[\angle DEC=\angle FEB\] And \[EC=EB\]. \[\therefore \]\[\angle EDC\cong \Delta EFB\] and therefore. \[BF=DC\] \[\therefore \] \[AF=\left( AB+BF \right)=\left( AB+DC \right)=2AB\].
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