9.01 Angle bisectors and perpendicular bisectors
Angle bisectors
Recall that an angle bisector is a ray from the vertex of the angle that divides an angle into two congruent angles.
Exploration
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?
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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 angle-angle-side 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. |
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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 |
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We can prove both the theorem and its converse using congruent triangles.
Perpendicular bisectors
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.
Exploration
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?
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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. |
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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 |
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The converse of the theorem is also true, and we can prove it using congruent triangles as well.