Recall that an angle bisector is a ray from the vertex of the angle that divides an angle into two congruent angles.
Let's have a look at the applet below, which shows the constructed angle bisector of $\angle ABC$∠ABC. If we move point $P$P along the angle bisector, or change the size of $\angle ABC$∠ABC, what relationship is always true? Can we explain why?

From the applet we can see that point P is always the same distance from the sides of the angle, no matter where it is on the angle bisector or how large we make $\angle ABC$∠ABC.
Why does this happen? In the applet we can see two right triangles formed. We can prove these triangles are congruent by angleangleside congruence. Therefore, the distances, being parts of those triangles, are also always the same.
Angle bisector theorem  If a point is on the bisector of an angle, then it is equidistant from the sides of the angle.
If  $\overrightarrow{AD}$›‹AD bisects $\angle BAC$∠BAC and $\overline{DB}\perp\overline{AB}$DB⊥AB and $\overline{DC}\perp\overline{AC}$DC⊥AC  then  $DB=DC$DB=DC. 
Converse of the angle bisector theorem  If a point in the interior of an angle is equidistant from the sides of the angle, then it is on the bisector of the angle.
If  $BD=DC$BD=DC and $\overline{DB}\perp\overline{AB}$DB⊥AB and $\overline{DC}\perp\overline{AC}$DC⊥AC  then  $\overrightarrow{AD}$›‹AD bisects $\angle BAC$∠BAC 
We can prove both the theorem and its converse using congruent triangles.
Recall that a segment bisector is a segment, line, or plane that intersects a segment at its midpoint. If a bisector is also perpendicular to the segment, it is called a perpendicular bisector.
Let's have a look at the properties of a perpendicular bisector. Move points, $A$A, $B$B, and $P$P in the applet below. What relationships in the diagram are always true? Can you explain why?

The perpendicular bisector of a line segment is the set of all points that are equidistant from the segment's endpoints. Notice, that no matter where we move point $P$P, it is always the same distance from points $A$A and $B$B. This is a theorem that can be proven using congruent triangles.
Perpendicular bisector theorem  In a plane, if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
If  $\overleftrightarrow{CP}$›‹CP is the $\perp$⊥ bisector of $\overline{AB}$AB  then  $CA=CB$CA=CB. 
Converse of the perpendicular bisector theorem  In a plane, if a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.
If  $DA=DB$DA=DB  then  $D$D lies on the $\perp$⊥ bisector of $\overline{AB}$AB 
The converse of the theorem is also true, and we can prove it using congruent triangles as well.