Question

(b) Show that the three bisectors of the an- gles of a triangle are concurrent, that is,they all pass through some point P.

Answer 1

Proof Figure 1 shows the triangle ABC with the angle bisectors AD,            BE and CF of its three angles A, B and C respectively. The points D, E and F are the intersection points of the angle bisectors and the opposite triangle sides. Since the straight lines AD and BE are the angle bisectors to the angles A and B respectively, they can not be parallel, otherwise the sides ABand BC would be in one straight line what is not the case. Therefore, the straight lines AD and BE intersect in some point P. From the lesson An angle bisector properties (Theorem 1) we know that the points of the angle bisector AD are equidistant from the sides AB and AC of the angle BAC.

Figure 1. To the Theorem

Figure 2. To the proof of the Theorem

In particular, the point P is equidistant from the sides AB and AC of the angle BAC. This means that the perpendiculars GP and HP (Figure 2) drawn from the point P to
the sides AB and AC are of equal length: GP = HP.

By the same reason, the points of the angle bisector BE are equidistant from the sides AB and BC of the angle ABC. In particular, the point P is equidistant from
the sides AB and BC of the angle ABC. This means that the perpendiculars GP and IP (Figure 2) drawn from the point P to the sides AB and BC are of equal length: GP = IP.

Two equalities above imply that the perpendiculars HP and IP are of equal length too: HP = IP. In other words, the point P is equidistant from the sides AC and BC of the
angle ACB. In turn, it implies that the intersection point P lies at the angle bisector CF of the angle ACB in accordance to the Theorem 2 of the lesson
An angle bisector properties. In other words, the angle bisector CF of the angle ACB passes through the point P.