This is based on the property of angle bisectors.  Given an angle and its bisector, any point on the angle bisector has the same distance to the two sides of the angle.  Conversely, any point with equal distance to the two sides must also be on the angle bisector.

For this problem, it is usually easier to take two of the angle bisectors in the triangle, and look at their intersection point, using the property above, and show that this intersection point is also on the third angle bisector.