In triangle ABC the internal bisector of ∠A is produced to meet the circumcircle a second time at M, and I is the incentre of the triangle. Which statement is correct?

The bisector of an inscribed angle splits the opposite arc in half, so M is the midpoint of arc BC and the equal arcs give MB = MC. For the third equality, chase the angles in triangle MBI: ∠MBI = ∠MBC + ∠CBI = ∠A/2 + ∠B/2, since ∠MBC stands on the same arc MC as ∠MAC = ∠A/2. Meanwhile the exterior angle ∠MIB of triangle ABI equals ∠IAB + ∠IBA = ∠A/2 + ∠B/2 as well. Two equal angles make triangle MBI isosceles, so MB = MI — the incentre–excentre lemma. MA is longer than MB except when AB = AC, and M lies ON the circle, not on the side BC; the point opposite A is a different point unless the triangle is isosceles at A.