Halving an angle without a protractor
The bisector of an angle is the line from the vertex that splits the angle into two equal angles. With a protractor you could measure the angle and halve the number, but only as closely as the scale can be read. A compass and a straightedge halve an angle exactly, whatever its size, and nothing but the two arms is needed.
Step 1: one arc across both arms
Put the compass point on the vertex V and draw one arc that cuts both arms. Call the two cuts X and Y. Both are one compass width from V, so VX = VY: the arc has marked the two arms at the same distance out.
One arc from V cuts both arms. The dashed line is the compass width, the same to each cut.
Step 2: equal arcs from the two cuts
Put the compass point on X and draw an arc inside the angle. Keep the same width, put the point on Y, and draw a second arc that crosses the first. Call the crossing Z.
Z is on both arcs, so it is the same distance from X as it is from Y. This width does not have to match the first arc’s. It only has to be the same for both arcs, and wide enough for them to meet.
Equal arcs from the two cuts meet inside the angle. The dashed lines from the two cuts to that point are equal.
Step 3: rule from the vertex
Rule a line from V through Z. It splits the angle into two equal halves, ∠XVZ = ∠ZVY. Measure to check: a 70° angle should come out as 35° and 35°.
The ruled line from V splits the angle into two halves. The tick on each half marks them as equal.
Why the two halves are equal
Look at the two cuts, X and Y. V is the same distance from both of them, because of Step 1. Z is the same distance from both of them, because of Step 2. The points that are equally far from X and Y are exactly the points on the perpendicular bisector of XY, so V and Z both lie on it, and the line VZ is the perpendicular bisector of XY.
Now fold the page along VZ. The fold line crosses XY at its midpoint and at a right angle, so the fold carries X exactly onto Y. V is on the fold, so it stays where it is. The arm through X therefore lands on the arm through Y, and the angle on one side of the fold lands exactly on the angle on the other side. The two halves match, so they are equal: VZ is a line of symmetry of the angle.
Any angle at all
None of the three steps used the size of the angle. A wide angle is bisected by the same moves, and so is an obtuse one: bisecting 140° gives 70° and 70°.
Even a straight angle works. Bisect the 180° angle at a point on a straight line and the two halves are 90° each, so the same moves construct the line at a right angle to a straight line through a point on it.
The same three steps on a wider angle. The bisector again splits it into two equal halves.
Three common mistakes
Using two different widths for the arcs from X and Y. They still meet, but at a point nearer one arm than the other, so the line leans off the middle.
Joining X and Y. That line runs across the angle from arm to arm. A bisector has to start at the vertex.
Halving the length VZ. Halving a length is not halving an angle: the bisector is a direction from V, and how far it runs does not matter.
Worked example: Bisectors of Two Angles on a Straight Line
Question AOB is a straight line and ∠ AOC = 70°. OD bisects ∠ COB and OE bisects ∠ AOC. Find ∠ DOE.
1.AOB is a straight line, so ∠ AOC + ∠ COB = 180° and ∠ COB = 180° − 70° = 110°.
On the straight line AOB, ∠ COB = 180° − 70° = 110°. 2.OD bisects ∠ COB: ∠ COD = 110° ÷ 2 = 55°.
OD bisects ∠ COB: ∠ COD = 55°. 3.OE bisects ∠ AOC: ∠ EOC = 70° ÷ 2 = 35°.
OE bisects ∠ AOC: ∠ EOC = 35°. 4.∠ DOE is the two halves next to OC: ∠ EOC + ∠ COD = 35° + 55° = 90°.
∠ DOE = 35° + 55° = 90°. 5.In general: half of ∠ AOC plus half of ∠ COB is half of 180°.
Half of ∠ AOC plus half of ∠ COB is half of 180°: the bisectors are always perpendicular. 6.So the bisectors of two angles on a straight line are always perpendicular, whatever ∠ AOC is.
Answer: ∠ DOE = 90°
Common mistakes
- Halving 70° and stopping: ∠ DOE needs the half of ∠ COB as well.
- Drawing OD inside ∠ AOC: OD bisects ∠ COB, on the other side of OC from OE.