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 $\triangle abc$.

statement
$q$ is the angle bisector of $angle a$ and $r$ is the angle bisector of $angle b$.
$p$ is the point of intersection of $q$ and $r$.
construct $overline{mp}$ such that and $m$ is on $overline{ab}$ and $overline{op}$ such that and $o$ is on $overline{ac}$.

reason
given
$q$ and $r$ are not parallel, so they intersect.
there exists a unique line perpendicular to a line, containing a point not on the line.

Explanation:

Step1: Angle - bisector property

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's construct \(PM\perp AB\) (\(\overline{MP}\perp\overline{AB}\)) and \(PN\perp BC\) (\(\overline{NP}\perp\overline{BC}\)) and \(PO\perp AC\) (\(\overline{OP}\perp\overline{AC}\)).
Because \(P\) is on \(q\) (the angle bisector of \(\angle A\)), \(PM = PO\). Because \(P\) is on \(r\) (the angle bisector of \(\angle B\)), \(PM=PN\).

Step2: Transitive property

From \(PM = PO\) and \(PM = PN\), by the transitive property of equality, \(PO=PN\).
Since \(P\) is equidistant from \(AB\) and \(AC\) (because \(PM = PO\)) and \(P\) is equidistant from \(AB\) and \(BC\) (because \(PM = PN\)), and \(s\) is the angle bisector of \(\angle C\). A point equidistant from two sides of an angle lies on the angle bisector of the angle. So \(P\) lies on \(s\) (the angle bisector of \(\angle C\)).

Answer:

Since \(P\) is on the angle bisector \(q\) of \(\angle A\), \(P\) is equidistant from \(AB\) and \(AC\). Since \(P\) is on the angle bisector \(r\) of \(\angle B\), \(P\) is equidistant from \(AB\) and \(BC\). By the transitive property, \(P\) is equidistant from \(AC\) and \(BC\), so \(P\) lies on the angle bisector \(s\) of \(\angle C\). Thus, \(q\), \(r\), and \(s\) are concurrent at \(P\) and \(P\) is equidistant from the sides of \(\triangle ABC\).