Ninth Grade Mathematics: Perpendicular Bisectors & Angle Bisectors
The two most fundamental building blocks of all geometric constructions are the perpendicular bisector of a line segment and the bisector of an angle. In this chapter, you will master the precise compass-and-straightedge protocols for both constructions and examine the rigorous congruence proofs that guarantee their mathematical precision.
Constructing the Perpendicular Bisector of a Segment
The perpendicular bisector of a line segment AB is the unique straight line that passes through the midpoint of AB at a right angle (90 degrees).
Construction Algorithm for Perpendicular Bisector:
1. Given line segment AB.
2. Place the compass point at A. Open the compass to a radius strictly
GREATER THAN HALF of AB.
3. Draw two arcs, one above the segment AB and one below the segment AB.
4. Keeping the exact same compass radius, place the compass point at B.
5. Draw two arcs intersecting the previous arcs at points P (above) and Q (below).
6. Using a straightedge, draw the straight line connecting P and Q.
P (Upper Arc Intersection)
*
/|\
/ | \
/ | \
A *----+---M---+----* B
\ | /
\ | /
\|/
*
Q (Lower Arc Intersection)
Line PQ is the Perpendicular Bisector of AB!
Line PQ intersects AB at point M. Point M is the exact midpoint of AB (AM = MB), and line PQ is perpendicular to AB at 90 degrees.
Proof of the Construction
Why is line PQ guaranteed to be the perpendicular bisector?
Proof:
Step 1: Join AP, BP, AQ, and BQ.
Step 2: By construction, all four arcs were drawn with the exact same radius r:
AP = BP = AQ = BQ = r.
Step 3: Quadrilateral APBQ has four equal sides, making it a RHOMBUS!
Step 4: In any rhombus, the diagonals are perpendicular bisectors of each other:
- Diagonal PQ bisects diagonal AB: AM = MB.
- Diagonal PQ is perpendicular to AB: Angle PMA = Angle PMB = 90 degrees.
The construction is proven. Every point lying on line PQ is equidistant from endpoints A and B.
Constructing the Bisector of an Angle
An angle bisector is a ray that originates from the vertex of an angle and divides the interior opening into two equal angles.
Construction Algorithm for Angle Bisector:
1. Given Angle ABC with vertex B.
2. Place the compass point at vertex B. With any convenient radius,
draw an arc intersecting ray BA at point P and ray BC at point Q.
3. Place the compass point at P. With a radius greater than half of PQ,
draw an arc in the interior of the angle.
4. With the exact same radius, place the compass point at Q and draw an arc
intersecting the previous arc at point R.
5. Draw the ray originating at vertex B and passing through R: Ray BR.
A
^
/
P * (Arc intersection)
/ \
/ \
/ * R (Crossing Arcs)
/ /
*-----+--------------------> C
B Q
Ray BR Bisects Angle ABC!
Ray BR is the exact angle bisector: Angle ABR = Angle CBR.
Proof of the Angle Bisector Construction
We prove this construction using the SSS Congruence Criterion:
Proof:
Step 1: Join PR and QR with straight segments.
Step 2: In Triangle BPR and Triangle BQR:
- BP = BQ (Radii of the initial circle drawn from center B)
- PR = QR (Arcs drawn with identical radius from P and Q)
- BR = BR (Common Side)
Step 3: By the SSS Congruence Criterion:
Triangle BPR =~ Triangle BQR.
Step 4: By CPCTC:
Angle PBR = Angle QBR.
Since Angle PBR = Angle ABR and Angle QBR = Angle CBR, ray BR bisects Angle ABC into two equal halves.
Chapter Exercises
Exercise 1: Describe the step-by-step procedure to find the exact center of a circle drawn on a piece of paper using only an unmarked straightedge and a compass.
Exercise 2: Explain why, in the perpendicular bisector construction, the compass opening must be strictly greater than half the length of the segment. What happens if the radius is less than half?
Exercise 3: Construct an angle of 45 degrees by first constructing a right angle (90 degrees) and then bisecting it. Describe all construction steps in clear paragraph form.
Solutions and Explanations
Solution 1: To locate the unknown center of a circle: draw any two non-parallel chords, say AB and CD, on the circle using a straightedge. Construct the perpendicular bisector of chord AB using intersecting compass arcs. Next, construct the perpendicular bisector of chord CD using the same protocol. By the Perpendicular Chord Bisection Theorem, the perpendicular bisector of any chord must pass directly through the center of the circle. Therefore, the unique point where the two perpendicular bisector lines intersect is the exact center O of the circle.
Solution 2: In the perpendicular bisector construction, the compass radius r represents the distance from endpoints A and B to the intersection points P and Q. If the radius r were chosen to be strictly less than half the length of AB (r < AB/2), the two circular arcs centered at A and B would have a combined reach of 2*r < AB. Because their distance from each other is AB, the arcs would never meet or intersect in the plane. If r = AB/2, the arcs would touch at only a single point lying directly on segment AB, failing to generate the two distinct points P and Q needed to determine a line. Choosing r > AB/2 guarantees that the two circles intersect at two distinct points on opposite sides of the segment, allowing the perpendicular bisector to be drawn.
Solution 3: To construct a 45-degree angle: begin by drawing a straight horizontal line segment AB and extending it past A to a point C. Construct a perpendicular line at vertex A by drawing arcs from A cutting the line at two points, and then drawing intersecting arcs of larger radius above the line; drawing the vertical ray through this intersection erects a 90-degree right angle (Angle DAB = 90 degrees). Now, bisect this 90-degree angle using the angle bisector construction: place the compass point at A, draw an arc intersecting ray AB at point P and the vertical ray AD at point Q. From points P and Q, draw intersecting arcs of equal radius in the interior of the angle, meeting at point R. Drawing ray AR from vertex A through R divides the 90-degree angle into two equal halves: 90 / 2 = 45 degrees. Thus, Angle RAB is an exact 45-degree angle.