The Central Idea
A bisector divides something into two equal parts.
For a segment, equal division concerns length:
\[ AM=MB. \]For an angle, equal division concerns angle measure:
\[ \angle ABD\cong\angle DBC. \]These equalities are not visual estimates. They are exact geometric conditions.
Two Different Bisector Concepts
The word bisector appears in two closely related but distinct settings.
- A segment bisector passes through the midpoint of a segment.
- An angle bisector divides an angle into two congruent angles.
A perpendicular bisector adds a second requirement to segment bisection: perpendicularity.
What Is a Segment Bisector?
A segment bisector is any line, ray, segment, or other suitable geometric object that passes through the midpoint of a segment.
If \(M\) is the midpoint of \(\overline{AB}\), then
\[ AM=MB. \]A segment bisector does not have to be perpendicular to the segment.
What Is a Perpendicular Bisector?
The perpendicular bisector of \(\overline{AB}\) is the line that passes through the midpoint of \(\overline{AB}\) and is perpendicular to the segment.
If it meets \(\overline{AB}\) at \(M\), then
\[ AM=MB \]and
\[ \ell\perp\overline{AB}. \]Two Conditions Are Required
To establish that a line is the perpendicular bisector of a segment, both conditions must be satisfied.
- The line passes through the midpoint.
- The line is perpendicular to the segment.
Either condition alone is insufficient.
A Perpendicular Line Need Not Bisect
A line can cross \(\overline{AB}\) at a right angle without passing through its midpoint.
Then the line is perpendicular to the segment but is not its perpendicular bisector.
A Segment Bisector Need Not Be Perpendicular
A line can pass through the midpoint \(M\) of \(\overline{AB}\) at an oblique angle.
Then it bisects the segment because
\[ AM=MB, \]but it is not the perpendicular bisector.
Constructing the Perpendicular Bisector
Given segment \(\overline{AB}\):
- Open the compass to a radius greater than half of \(AB\).
- Center at \(A\) and draw arcs above and below the segment.
- Without changing the compass opening, center at \(B\) and draw arcs that intersect the first pair.
- Label the intersections \(P\) and \(Q\).
- Draw line \(\overleftrightarrow{PQ}\).
The line \(\overleftrightarrow{PQ}\) is the perpendicular bisector of \(\overline{AB}\).
Why Equal Compass Radii Matter
Since \(P\) lies on equal-radius arcs centered at \(A\) and \(B\),
\[ PA=PB. \]Likewise,
\[ QA=QB. \]Thus both constructed points are equidistant from the endpoints of the segment.
Why the Arc Radius Must Exceed Half the Segment
The circles centered at \(A\) and \(B\) must intersect in two points.
If the common compass radius is \(r\), then in the ordinary construction we need
\[ r>\frac{AB}{2}. \]Otherwise the circles either fail to intersect or meet only in a degenerate boundary case.
The Perpendicular-Bisector Theorem
Every point on the perpendicular bisector of a segment is equidistant from the segment's endpoints.
If \(P\) lies on the perpendicular bisector of \(\overline{AB}\), then
\[ \boxed{PA=PB}. \]Proof of the Perpendicular-Bisector Theorem
Let \(M\) be the midpoint of \(\overline{AB}\), and let \(P\) lie on the perpendicular bisector.
Then
\[ AM=MB, \] \[ \angle PMA=\angle PMB=90^\circ, \]and
\[ PM=PM. \]Therefore \(\triangle PMA\) and \(\triangle PMB\) are congruent, giving
\[ PA=PB. \]The Converse Perpendicular-Bisector Theorem
If a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of that segment.
If
\[ PA=PB, \]then \(P\) lies on the perpendicular bisector of \(\overline{AB}\).
The Perpendicular Bisector as a Locus
The theorem and its converse combine into a locus statement:
\[ \boxed{ \text{The perpendicular bisector of }\overline{AB} \text{ is the set of all points }P\text{ such that }PA=PB. } \]This is one of the most important locus relationships in elementary geometry.
Why the Locus View Is Powerful
A perpendicular bisector is not merely a line obtained from a construction.
It is the complete solution set to the condition
\[ PA=PB. \]This turns distance questions into geometric-location questions.
Example: Find Points Equidistant from Two Locations
Suppose \(A\) and \(B\) represent two facilities.
Every point that is the same straight-line distance from both facilities lies on the perpendicular bisector of \(\overline{AB}\).
The locus therefore divides the plane into points closer to \(A\), points closer to \(B\), and points equally distant from both.
The Perpendicular Bisector Divides the Plane
One side of the perpendicular bisector consists of points closer to \(A\).
The other side consists of points closer to \(B\).
The perpendicular bisector itself is the equality boundary:
\[ PA=PB. \]Constructing a Midpoint from the Perpendicular Bisector
If the perpendicular bisector of \(\overline{AB}\) intersects the segment at \(M\), then
\[ \boxed{AM=MB}. \]Thus the construction simultaneously creates the midpoint.
Midpoint Formula
In coordinates, if
\[ A(x_1,y_1) \]and
\[ B(x_2,y_2), \]then the midpoint is
\[ \boxed{ M\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right). } \]Slope of a Perpendicular Bisector
If \(\overline{AB}\) has finite nonzero slope \(m\), then its perpendicular bisector has slope
\[ -\frac{1}{m}. \]The perpendicular bisector also passes through the midpoint of \(\overline{AB}\).
Coordinate Example: Perpendicular Bisector
Let
\[ A(0,0),\qquad B(6,4). \]The midpoint is
\[ M(3,2). \]The slope of \(\overline{AB}\) is
\[ \frac{4}{6}=\frac23. \]Therefore the perpendicular slope is
\[ -\frac32. \]Equation of the Perpendicular Bisector
Using point-slope form through \(M(3,2)\):
\[ y-2=-\frac32(x-3). \]Every point on this line is equidistant from \(A\) and \(B\).
Distance-Formula Derivation
The same perpendicular bisector can be found directly from the locus condition.
Let \(P(x,y)\) satisfy
\[ PA=PB. \]Then
\[ x^2+y^2=(x-6)^2+(y-4)^2. \]Simplifying produces the equation of the perpendicular bisector.
Why Squared Distances Are Convenient
To solve
\[ PA=PB, \]we can equate squared distances:
\[ PA^2=PB^2. \]This avoids unnecessary square roots while preserving the equality because distances are nonnegative.
Horizontal Segment Case
Suppose
\[ A(-4,3),\qquad B(6,3). \]The midpoint is
\[ M(1,3). \]Since \(\overline{AB}\) is horizontal, its perpendicular bisector is vertical:
\[ \boxed{x=1}. \]Vertical Segment Case
Suppose
\[ A(2,-5),\qquad B(2,7). \]The midpoint is
\[ M(2,1). \]Since \(\overline{AB}\) is vertical, its perpendicular bisector is horizontal:
\[ \boxed{y=1}. \]Perpendicular Bisectors in Triangles
Every side of a triangle has a perpendicular bisector.
The three perpendicular bisectors meet at one common point.
This point is the circumcenter.
Why the Circumcenter Is Equidistant from the Vertices
If \(O\) lies on the perpendicular bisector of \(\overline{AB}\), then
\[ OA=OB. \]If \(O\) also lies on the perpendicular bisector of \(\overline{BC}\), then
\[ OB=OC. \]Therefore,
\[ \boxed{OA=OB=OC}. \]The Circumcircle
Because the circumcenter \(O\) is equidistant from all three vertices, a circle centered at \(O\) with radius
\[ OA \]passes through \(A\), \(B\), and \(C\).
This is the triangle's circumcircle.
Circumcenter Location
The circumcenter's location depends on the triangle.
- For an acute triangle, it lies inside.
- For a right triangle, it lies at the midpoint of the hypotenuse.
- For an obtuse triangle, it lies outside.
Perpendicular Bisectors and Voronoi Boundaries
In spatial partitioning, the perpendicular bisector between two sites forms the boundary separating points closer to one site from points closer to the other.
Larger systems of such boundaries create Voronoi diagrams.
The elementary locus principle scales into an important computational and applied geometric structure.
What Is an Angle Bisector?
An angle bisector is a ray from the vertex of an angle that divides the angle into two congruent angles.
If ray \(\overrightarrow{BD}\) bisects \(\angle ABC\), then
\[ \boxed{\angle ABD\cong\angle DBC}. \]Angle Measure and Bisection
If
\[ m\angle ABC=\theta, \]then an angle bisector produces
\[ m\angle ABD = m\angle DBC = \frac{\theta}{2}. \]Example: Bisecting a \(70^\circ\) Angle
If
\[ m\angle ABC=70^\circ, \]then its bisector creates two angles of
\[ \boxed{35^\circ}. \]Constructing an Angle Bisector
Given \(\angle ABC\):
- Center the compass at \(B\) and draw an arc crossing the two rays at \(E\) and \(F\).
- Using an equal compass radius, draw arcs centered at \(E\) and \(F\) that intersect inside the angle at \(D\).
- Draw ray \(\overrightarrow{BD}\).
The ray \(\overrightarrow{BD}\) is the angle bisector.
Why the First Arc Matters
The first arc is centered at \(B\), so
\[ BE=BF. \]This provides one pair of equal sides in the two triangles used to justify the construction.
Why the Second Pair of Arcs Matters
The equal-radius arcs centered at \(E\) and \(F\) intersect at \(D\), so
\[ ED=FD. \]The two small triangles also share
\[ BD=BD. \]Proof of the Angle-Bisector Construction
In \(\triangle BED\) and \(\triangle BFD\),
\[ BE=BF, \] \[ ED=FD, \]and
\[ BD=BD. \]Therefore the triangles are congruent by SSS.
Hence,
\[ \boxed{\angle EBD\cong\angle DBF}. \]Distance from a Point to a Line
The distance from a point to a line is measured along the perpendicular from the point to the line.
This distance concept is essential to the locus characterization of angle bisectors.
The Angle-Bisector Theorem for Distance to Sides
Every point on an angle bisector is equidistant from the two sides of the angle.
If \(P\) lies on the bisector of \(\angle ABC\), and perpendiculars from \(P\) to the sides meet them at \(E\) and \(F\), then
\[ \boxed{PE=PF}. \]Why the Distance-to-Sides Theorem Works
The perpendiculars create right triangles \(\triangle PBE\) and \(\triangle PBF\).
They share hypotenuse \(\overline{PB}\), and the angle at \(B\) is bisected.
The triangles are congruent, so the perpendicular distances satisfy
\[ PE=PF. \]The Converse Angle-Bisector Locus Theorem
If an interior point of an angle is equidistant from the two sides of the angle, then it lies on the angle bisector.
This gives the locus statement:
\[ \boxed{ \text{The angle bisector is the locus of interior points equidistant from the two sides.} } \]Internal and External Angle Bisectors
Two intersecting lines determine both internal and external angle-bisector directions.
An internal angle bisector divides the chosen interior angle.
An external angle bisector divides the supplementary exterior angle.
Internal and External Bisectors Are Perpendicular
If two adjacent angles are supplementary, their measures sum to
\[ 180^\circ. \]Their half-measures therefore sum to
\[ 90^\circ. \]Thus the internal and external angle bisectors are perpendicular.
Angle Bisectors in Triangles
Every angle of a triangle has an internal angle bisector.
The three internal angle bisectors are concurrent.
Their common point is the incenter.
Why the Incenter Is Equidistant from the Sides
Because the incenter lies on the bisector of \(\angle A\), it is equidistant from sides \(AB\) and \(AC\).
Because it also lies on another angle bisector, the corresponding side distances are equal.
Therefore the incenter is equally distant from all three sides.
The Incircle
Let \(I\) be the incenter, and let \(r\) be the perpendicular distance from \(I\) to any side.
A circle centered at \(I\) with radius \(r\) is tangent to all three sides.
This circle is the triangle's incircle.
Angle Bisector Theorem in a Triangle
A different theorem also carries the name Angle Bisector Theorem.
If \(AD\) bisects \(\angle A\) in \(\triangle ABC\) and \(D\) lies on \(\overline{BC}\), then
\[ \boxed{\frac{BD}{DC}=\frac{AB}{AC}}. \]What the Triangle Angle Bisector Theorem Means
An angle bisector does not generally divide the opposite side into two equal lengths.
Instead, it divides that side proportionally to the adjacent side lengths.
Equal division occurs only in special cases such as an isosceles triangle with
\[ AB=AC. \]Example: Using the Triangle Angle Bisector Theorem
Suppose
\[ AB=6,\qquad AC=9, \]and \(AD\) bisects \(\angle A\).
Then
\[ \frac{BD}{DC} = \frac{6}{9} = \frac23. \]The opposite side is divided in a \(2:3\) ratio.
Finding the Two Parts of the Opposite Side
Suppose additionally
\[ BC=20. \]Since
\[ BD:DC=2:3, \]there are \(5\) equal ratio parts, each of length
\[ \frac{20}{5}=4. \]Therefore,
\[ \boxed{BD=8,\qquad DC=12}. \]Converse of the Triangle Angle Bisector Theorem
If \(D\) lies on \(\overline{BC}\) and
\[ \frac{BD}{DC}=\frac{AB}{AC}, \]then \(\overline{AD}\) bisects \(\angle A\).
This converse can be useful in proof.
Isosceles Triangle Connection
In an isosceles triangle with
\[ AB=AC, \]the angle bisector from vertex \(A\) also divides the base equally:
\[ BD=DC. \]In this special situation, the same segment is an angle bisector, median, altitude, and perpendicular bisector of the base.
One Segment, Several Roles
In an isosceles triangle, symmetry causes several geometric constructions to coincide.
From the apex to the base, the symmetry line is simultaneously:
- an angle bisector;
- a median;
- an altitude;
- a perpendicular bisector of the base.
Do Not Assume These Roles Coincide in Every Triangle
In a general scalene triangle, the angle bisector, median, altitude, and perpendicular bisector are different lines or segments.
Their coincidence is a special consequence of symmetry, not a universal triangle property.
Coordinate Angle Bisectors
Angle bisectors can also be described algebraically.
For two intersecting lines, the angle bisectors consist of points whose perpendicular distances to the two lines are equal.
This converts the locus condition into an equation.
Distance from a Point to a Line
For line
\[ Ax+By+C=0, \]the perpendicular distance from point \((x_0,y_0)\) is
\[ \boxed{ d= \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}. } \]Equation of Two Line Bisectors
For lines
\[ A_1x+B_1y+C_1=0 \]and
\[ A_2x+B_2y+C_2=0, \]their angle bisectors satisfy
\[ \boxed{ \frac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}} = \pm \frac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}. } \]Why the Plus-or-Minus Appears
Two intersecting lines have two angle-bisector lines.
One bisects one pair of vertical angles, and the other bisects the supplementary pair.
The two signs produce these two loci.
Simple Coordinate Example
Consider the coordinate axes:
\[ x=0 \]and
\[ y=0. \]Points equidistant from the axes satisfy
\[ |x|=|y|. \]Therefore the two angle bisectors are
\[ \boxed{y=x} \]and
\[ \boxed{y=-x}. \]Angle Bisectors and Reflection
Reflection across an angle bisector maps one side of the angle onto the other.
This transformation viewpoint explains both equal angle measures and equal perpendicular distances to the sides.
Perpendicular Bisectors and Reflection
Reflection across the perpendicular bisector of \(\overline{AB}\) maps endpoint \(A\) to endpoint \(B\).
Every point on the reflection axis remains fixed, which explains why such points have equal distances to \(A\) and \(B\).
Bisectors as Symmetry Axes
Both major bisector types can therefore be interpreted through symmetry.
- The perpendicular bisector exchanges two segment endpoints.
- The angle bisector exchanges the two sides of the angle under reflection.
Perpendicular Bisectors and Circles
The perpendicular bisector of a chord passes through the center of the circle.
If chord endpoints are \(A\) and \(B\), the center \(O\) satisfies
\[ OA=OB \]because both are radii.
Therefore \(O\) lies on the perpendicular bisector of \(\overline{AB}\).
Finding the Center of a Circle
If the center of a drawn circle is unknown:
- draw one chord and construct its perpendicular bisector;
- draw a second nonparallel chord and construct its perpendicular bisector;
- locate their intersection.
That intersection is the circle's center.
Angle Bisectors and Tangent Circles
A circle tangent to both sides of an angle must have its center equidistant from the two sides.
Therefore its center lies on an angle bisector.
This is the fundamental reason the incenter lies on all three angle bisectors of a triangle.
Constructing a Circle Tangent to Both Sides of an Angle
Choose a center \(P\) on the angle bisector.
Drop a perpendicular from \(P\) to either side of the angle. Let that distance be \(r\).
A circle centered at \(P\) with radius \(r\) is tangent to both sides.
Bisectors as Constraint Solvers
A perpendicular bisector solves
\[ PA=PB. \]An angle bisector solves
\[ d(P,\ell_1)=d(P,\ell_2). \]These equations express the two constructions as spatial constraint systems.
Combining Bisector Loci
Many geometry problems ask for a point satisfying several equal-distance conditions at once.
The solution is often an intersection of bisector loci.
For example, a point equidistant from three triangle vertices lies at the intersection of perpendicular bisectors.
Combining Angle-Bisector Loci
A point equidistant from all three sides of a triangle lies on the angle bisectors of the triangle.
Their intersection is the incenter.
Bisectors in Construction Problems
Bisectors are frequently intermediate tools rather than the final requested objects.
They can help construct:
- circle centers;
- tangent circles;
- parallel lines;
- regular polygons;
- triangle centers;
- symmetry axes.
Perpendicular Bisector as a Decision Boundary
If two sites \(A\) and \(B\) compete by nearest distance, the perpendicular bisector is the boundary where neither is closer.
On the boundary,
\[ PA=PB. \]Crossing the boundary changes which site is closer.
Angle Bisector as a Directional Boundary
An angle bisector gives a direction exactly halfway between two rays in angular measure.
This makes angle bisectors useful in reflection, symmetry, navigation, and geometric optimization.
Common Error: Confusing Segment Bisector and Perpendicular Bisector
A segment bisector must pass through the midpoint.
A perpendicular bisector must also meet the segment at a right angle.
Do not add perpendicularity unless it is given or established.
Common Error: Assuming a Perpendicular Is a Bisector
A line crossing a segment at \(90^\circ\) is not necessarily its perpendicular bisector.
The intersection must also be the midpoint.
Common Error: Assuming a Midpoint Line Is Perpendicular
A line through a midpoint can meet the segment at many possible angles.
Midpoint incidence alone gives bisection, not perpendicularity.
Common Error: Estimating the Midpoint
A point that looks halfway along a segment is not automatically the midpoint.
Midpoint status requires
\[ AM=MB. \]Common Error: Estimating the Angle Bisector
A ray that appears centered inside an angle is not automatically the angle bisector.
The two resulting angle measures must be established as equal.
Common Error: Changing the Compass Radius
Equal-radius arcs are essential to the standard compass constructions.
Accidentally changing the compass width can destroy the equal-distance relationships used in the proof.
Common Error: Using an Arc Radius That Is Too Small
In a perpendicular-bisector construction, arcs centered at the endpoints must intersect.
The compass radius therefore needs to exceed half the segment length.
Common Error: Assuming an Angle Bisector Bisects the Opposite Side
In a general triangle, an angle bisector divides the opposite side in the ratio of the adjacent sides:
\[ \frac{BD}{DC} = \frac{AB}{AC}. \]It does not generally create
\[ BD=DC. \]Common Error: Confusing Distance to a Side with Distance to a Vertex
The angle-bisector locus theorem concerns perpendicular distance to the two sides of the angle.
It does not say that a point on the angle bisector is equally distant from arbitrary points chosen on those sides.
Common Error: Using Ordinary Distance Instead of Perpendicular Distance
Distance from a point to a line means the shortest distance, measured along a perpendicular.
An arbitrary slanted segment from the point to the line does not represent point-to-line distance.
Common Error: Forgetting the External Angle Bisector
Two intersecting lines have two angle-bisector lines.
One corresponds to the chosen interior pair of vertical angles and the other to the supplementary pair.
Common Error: Assuming Triangle Centers Are Always Inside
The incenter is always inside a nondegenerate triangle.
The circumcenter is not.
Its location changes with the triangle's angle type.
A Reliable Procedure for Perpendicular-Bisector Problems
- Identify the segment endpoints.
- Determine whether the goal concerns equal endpoint distances, a midpoint, or perpendicularity.
- Use equal-radius arcs or coordinate methods to locate the bisector.
- Translate the result into \(PA=PB\) when using the locus theorem.
- Verify both midpoint and perpendicular conditions when claiming a perpendicular bisector.
A Reliable Procedure for Angle-Bisector Problems
- Identify the angle and its vertex.
- Determine whether the goal concerns equal angles, equal distances to sides, or a triangle-side ratio.
- Use the correct angle-bisector theorem.
- Construct or identify the bisector.
- Translate the geometric condition into angle equalities, distance equalities, or proportions.
- Check that the relevant distances to sides are perpendicular distances.
Example: Perpendicular-Bisector Locus
Points \(A\) and \(B\) are fixed. Find all points \(P\) satisfying
\[ PA=PB. \]The complete solution is
\[ \boxed{\text{the perpendicular bisector of }\overline{AB}}. \]Example: Angle-Bisector Locus
Find all interior points \(P\) of \(\angle ABC\) that are equally distant from rays \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\).
The complete solution is
\[ \boxed{\text{the internal angle bisector of }\angle ABC}. \]Example: Circumcenter Construction
Given \(\triangle ABC\):
- construct the perpendicular bisector of \(\overline{AB}\);
- construct the perpendicular bisector of \(\overline{BC}\);
- let them intersect at \(O\).
Then
\[ \boxed{OA=OB=OC}. \]Example: Incenter Construction
Given \(\triangle ABC\):
- construct the bisector of \(\angle A\);
- construct the bisector of \(\angle B\);
- let them intersect at \(I\).
The third angle bisector also passes through \(I\).
Point \(I\) is equidistant from all three sides.
Example: Solve a Side-Ratio Problem
In \(\triangle ABC\), \(AD\) bisects \(\angle A\),
\[ AB=10,\qquad AC=15,\qquad BC=25. \]Then
\[ \frac{BD}{DC} = \frac{10}{15} = \frac23. \]Since \(BD+DC=25\), the five ratio parts each equal \(5\).
Therefore,
\[ \boxed{BD=10,\qquad DC=15}. \]Practice 1: Perpendicular-Bisector Conditions
Line \(\ell\) is the perpendicular bisector of \(\overline{AB}\), meeting it at \(M\).
State the two defining relationships.
Practice 2: Locus of Equal Endpoint Distance
What is the locus of all points \(P\) satisfying
\[ PA=PB? \]Practice 3: Midpoint
If \(M\) is the midpoint of \(\overline{AB}\) and
\[ AB=18, \]find \(AM\) and \(MB\).
Practice 4: Angle Bisection
Ray \(\overrightarrow{BD}\) bisects an angle measuring
\[ 96^\circ. \]Find the measures of the two smaller angles.
Practice 5: Angle-Bisector Locus
Point \(P\) lies inside an angle and is the same perpendicular distance from both sides.
What can be concluded about \(P\)?
Practice 6: Circumcenter
Point \(O\) lies on the perpendicular bisectors of \(\overline{AB}\) and \(\overline{BC}\).
State the relationship among \(OA\), \(OB\), and \(OC\).
Practice 7: Incenter
Point \(I\) is the intersection of two internal angle bisectors of a triangle.
What distance relationship holds between \(I\) and the triangle's three sides?
Practice 8: Triangle Angle Bisector Theorem
In \(\triangle ABC\), \(AD\) bisects \(\angle A\),
\[ AB=8,\qquad AC=12. \]Find the ratio
\[ BD:DC. \]Practice 9: Coordinate Perpendicular Bisector
Find the perpendicular bisector of the segment joining
\[ A(0,2) \]and
\[ B(6,2). \]Practice 10: Identify the Error
A student says, “This line passes through the midpoint of \(\overline{AB}\), so it must be the perpendicular bisector.”
Explain the missing condition.
Practice Answers
Practice 1
The line passes through the midpoint and is perpendicular to the segment:
\[ \boxed{AM=MB} \]and
\[ \boxed{\ell\perp\overline{AB}}. \]Practice 2
The locus is
\[ \boxed{\text{the perpendicular bisector of }\overline{AB}}. \]Practice 3
A midpoint divides the segment into two equal lengths:
\[ AM=MB=\frac{18}{2}=9. \]Therefore,
\[ \boxed{AM=MB=9}. \]Practice 4
An angle bisector divides the angle into two equal measures:
\[ \frac{96^\circ}{2}=48^\circ. \]Therefore both smaller angles measure
\[ \boxed{48^\circ}. \]Practice 5
By the converse angle-bisector locus theorem, \(P\) lies on the internal angle bisector.
\[ \boxed{P\text{ lies on the angle bisector}.} \]Practice 6
From the first perpendicular bisector,
\[ OA=OB. \]From the second,
\[ OB=OC. \]Therefore,
\[ \boxed{OA=OB=OC}. \]Practice 7
The incenter is equidistant from all three sides of the triangle, where distance means perpendicular distance.
\[ \boxed{d(I,AB)=d(I,BC)=d(I,CA)}. \]Practice 8
By the Triangle Angle Bisector Theorem,
\[ \frac{BD}{DC} = \frac{AB}{AC} = \frac{8}{12} = \frac23. \]Therefore,
\[ \boxed{BD:DC=2:3}. \]Practice 9
The midpoint is
\[ M(3,2). \]The original segment is horizontal, so its perpendicular bisector is the vertical line
\[ \boxed{x=3}. \]Practice 10
Passing through the midpoint establishes segment bisection, but not perpendicularity.
The missing condition is
\[ \boxed{\ell\perp\overline{AB}}. \]What to Remember
Perpendicular bisectors and angle bisectors both divide geometric structure evenly, but they encode different equality conditions.
- A perpendicular bisector passes through a segment's midpoint and meets the segment at \(90^\circ\).
- Its points are exactly the points equidistant from the segment endpoints.
- Triangle perpendicular bisectors meet at the circumcenter.
- An angle bisector divides an angle into two congruent angles.
- Its interior points are exactly the points equidistant from the two sides of the angle.
- Triangle angle bisectors meet at the incenter.
- The Triangle Angle Bisector Theorem divides the opposite side proportionally to the adjacent sides.
- Bisectors can be understood through construction, loci, coordinates, congruence, and reflection symmetry.
The most important distinction is: a bisector is defined by exact equality conditions, not by appearing centered in a drawing.
Where We Go Next
Perpendicular and angle bisectors give us reliable tools for creating equal distances, equal angles, right angles, centers, and symmetry axes.
The next lesson, Constructing Parallels, Tangents, and Regular Figures, uses these tools to build more complex Euclidean structures, including parallel lines, circle tangents, and exact regular polygons.