I decided to generalize a bit and let ABCDEF be six consecutive vertices of a regular polygon of N sides. Let the intersection of AD and CF be J.
Each side of the polygon has an associated arc whose arc length is 2pi/N radians.
Arc AC and DF each subtend two of those arcs, so the are each 4pi/N radians.
Angles DAF and CFA have a measure of half the subtended arcs DF and AC, so angles DAF and CFA are each 2pi/N radians.
Then triangle AJF has two angles of 2pi/N radians each. The remaining angle, AJF is then (1-4/N)*pi radians.
The supplementary angle of AJF may also be the smallest. That angle is 4pi/N radians.
For the specific problem we want the smaller of 4pi/N verses (1-4/N)*pi where N=7. 4pi/7 > (1-4/7)*pi = 3pi/7, so the measure of the smaller angle is 3pi/7 radians.
Solution
