如图,五边形ABCDE是⊙O的内接正五边形,AF是⊙O的直径,则∠BDF的度数是( )
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解:AF是圆O的直径,五边形ABCDE是圆O的内接正五边形, $\therefore \widehat{C F}=\widehat{D F},\widehat{B C}=\widehat{D E},\angle B A E=108^{\circ}$,
$\therefore \widehat{B F}=\widehat{E F}$,
$\therefore \angle B A F=\frac{1}{2} \angle B A E=54^{\circ}$,
$\therefore \angle B D F=\angle B A F=54^{\circ}$.