Question

given \\( \triangle abc \\) with angle bisector \\( q \\) of \\( \angle a \\), angle bisector \\( r \\) of \\( \angle b \\), and angle bisector \\( s \\) of \\( \angle c \\), prove \\( q, r \\), and \\( s \\) are concurrent at point \\( p \\) equidistant from the sides of \\( \triangle abc \\).

Explanation:

Step1: Use the Angle - Bisector Theorem

Since \(q\) is the angle bisector of \(\angle A\) and \(r\) is the angle bisector of \(\angle B\), let \(P\) be the intersection of \(q\) and \(r\). By the property of the angle bisector, the distance from a point on an angle bisector to the two sides of the angle is equal. So, the distance from \(P\) to \(AB\) and \(AC\) (using angle bisector \(q\)) is equal, and the distance from \(P\) to \(AB\) and \(BC\) (using angle bisector \(r\)) is equal.

Step2: Consider the third angle bisector

Let \(s\) be the angle bisector of \(\angle C\). We know that the set of points equidistant from two sides of an angle lies on the angle bisector of that angle. Since \(P\) is equidistant from \(AB\) and \(AC\) (from \(q\)) and equidistant from \(AB\) and \(BC\) (from \(r\)), then \(P\) is equidistant from \(AC\) and \(BC\). By the converse of the angle - bisector theorem, \(P\) lies on the angle bisector \(s\) of \(\angle C\).

Answer:

The angle bisectors \(q\), \(r\), and \(s\) are concurrent at point \(P\) which is equidistant from the sides of \(\triangle ABC\) because of the properties of angle bisectors (a point on an angle bisector is equidistant from the sides of the angle) and their converse (a point equidistant from the sides of an angle lies on the angle bisector).