In triangle ABC the internal bisector of ∠A is drawn together with the bisectors of the exterior angles at B and at C. What is true of these three lines?

A point on the internal bisector of ∠A is equidistant from the lines AB and AC; a point on the external bisector at B is equidistant from the lines BA and BC. Where the first two meet, all three lines AB, BC and CA are the same distance away, so that point must lie on the third bisector too — the three are concurrent at the excentre opposite A. The circle centred there touches the side BC itself and the two other sides only after they are produced. The incentre needs all THREE internal bisectors, and the circumcentre comes from the perpendicular bisectors of the sides, not from angle bisectors at all.