Answer: C
If \(AB:ED = AC:EF\), then
\(\angle C = \angle F\) and \(\angle B = \angle D\).
\(\angle B = 180^\circ - (40^\circ + \angle A + \angle C)\)
= \(180^\circ - (40^\circ + 65^\circ) = 75^\circ\)
If \(AB:ED = AC:EF\), then
\(\angle C = \angle F\) and \(\angle B = \angle D\).
\(\angle B = 180^\circ - (40^\circ + \angle A + \angle C)\)
= \(180^\circ - (40^\circ + 65^\circ) = 75^\circ\)