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 \(\angle B\), and \(P\) is the intersection of \(q\) and \(r\). By the property of the angle - bisector, a point on an angle bisector is equidistant from the sides of the angle. Let \(PM\perp AB\) and \(PO\perp AC\) (constructed as per the requirement). Since \(P\) is on \(q\) (angle bisector of \(\angle A\)), \(PM = PO\). Let \(PN\perp BC\) (construct another perpendicular). Since \(P\) is on \(r\) (angle bisector of \(\angle B\)), \(PM=PN\).

Step2: Prove the concurrency of the third bisector

We have shown that \(PM = PO\) and \(PM = PN\), so \(PO=PN\). By the converse of the angle - bisector theorem, if a point is equidistant from the two sides of an angle, then it lies on the angle bisector of that angle. So \(P\) lies on the angle bisector \(s\) of \(\angle C\).

Answer:

The angle bisectors \(q\), \(r\), and \(s\) are concurrent at point \(P\) (the in - center of \(\triangle ABC\)) which is equidistant from the sides of \(\triangle ABC\) because a point on an angle bisector is equidistant from the sides of the angle and by using the transitive property of equality for the distances from \(P\) to the sides of the triangle.