You are given the cyclic quadrilateral ABCD. The angle bisector of the angle at A meets the circle at E, and the angle bisector at C meets the circle at F. Show that EF is a diameter of the circle

Answer 1

The following are arcs:

DE ≅ BE, DF ≅ BF subtended by equal angles

DE + DF ≅ BE + BF segment addition

EF ≅ FE substitution

QED

Answer 2

Arc BE = m ∠A; Arc BF = m ∠C; So, Arc EF = m ∠A + m ∠C = 180°. So, EF is diameter