Yanıt: $\boxed{C}$
$m\left ( \widehat{FBD} \right )=m\left ( \widehat{FDB} \right )$ ve $m\left ( \widehat{EBF} \right )=m\left ( \widehat{EFB} \right )=2.m\left ( \widehat{FBD} \right )$ dir. Buradan, $m\left ( \widehat{EFB} \right )=30^{\circ}$ dir. $ABFD$ deltoit olduğundan $m\left ( \widehat{DFA} \right )=m\left ( \widehat{BFA} \right )=75 ^{\circ}$ dir.