Constructions: Bisecting Angles and Lines
A construction uses only a straight-edge (ruler) and a pair of compasses to draw exact geometric figures. Today: the perpendicular bisector of a line, the angle bisector, and the perpendicular from a point to a line.
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You have a long stick. How could you find the exact midpoint without measuring? List two methods (one with string, one with folding, etc.). Now imagine you have only a ruler and compasses on paper — what would you do?
A construction draws an exact geometric figure using only:
- A straight-edge (ruler, but you don't measure lengths with it — just use the edge).
- A pair of compasses (for drawing circles and arcs).
Today's three constructions:
- Perpendicular bisector of a line segment (cuts the segment in half AT right angles).
- Angle bisector (cuts an angle exactly in half).
- Perpendicular from a point to a line (drops a vertical from a point onto a given line).
Constructions are exact: they don't depend on measuring with a marked ruler. Two thousand years ago, Greek mathematicians proved which shapes can be constructed this way (it turns out many can, but not all — you can't trisect a general angle with just these tools!).
Know
- The two construction tools (straight-edge + compass)
- Definition of perpendicular bisector
- Definition of angle bisector
Understand
- Why both arcs need the same radius
- Why the perpendicular bisector passes through the midpoint
- Why the angle bisector is unique
Can Do
- Describe each construction step-by-step
- Construct a perpendicular bisector
- Construct an angle bisector
- Drop a perpendicular from a point to a line
Given a segment $AB$. To find a line that cuts $AB$ exactly in half at a right angle:
- Place the compass point at $A$. Open it to MORE than half the length of $AB$.
- Draw a large arc that goes above AND below the segment.
- Without changing the radius, move the compass point to $B$.
- Draw a second arc that crosses the first arc above AND below.
- Use the straight-edge to join the two intersection points. This line is the perpendicular bisector.
The construction works because every point on the perpendicular bisector is the same distance from $A$ and $B$ — that's exactly the definition of a perpendicular bisector. The two arcs share the radius, so any intersection point is equidistant from both endpoints.
Book notes · Perpendicular bisector
- Compass on $A$, radius $>$ half of $AB$; arc above and below.
- Same radius from $B$; second arc above and below.
- Join the two crossings — that's the perpendicular bisector.
Given an angle $\angle ABC$ with vertex at $B$. To draw a ray from $B$ that cuts $\angle ABC$ exactly in half:
- Place the compass point at $B$. Choose any radius.
- Draw an arc that crosses BOTH sides of the angle. Label the crossings $P$ (on $BA$) and $Q$ (on $BC$).
- Without changing the radius, place the compass at $P$ and draw an arc inside the angle.
- Same radius from $Q$; draw another arc inside, crossing the previous one at a point $R$.
- Draw the ray from $B$ through $R$. This is the angle bisector — $\angle ABR = \angle RBC$.
Why it works: $BP = BQ$ (same arc), and $PR = QR$ (same arc again). So $\triangle BPR$ and $\triangle BQR$ are congruent (SSS). The matching angles $\angle PBR$ and $\angle QBR$ are equal, so $BR$ bisects the angle.
Book notes · Angle bisector
- Arc from vertex crosses both arms at $P, Q$.
- Equal arcs from $P$ and $Q$ meet at $R$ inside the angle.
- Ray from vertex through $R$ bisects the angle.
The angle bisector divides an angle into two angles that are equal in measure.
Given a line $\ell$ and a point $P$ NOT on $\ell$. To drop a perpendicular from $P$ onto $\ell$:
- Compass on $P$. Open wide enough so an arc crosses $\ell$ at two points; call them $X$ and $Y$.
- Same (or larger) radius from $X$: draw an arc on the opposite side of $\ell$ from $P$.
- Same radius from $Y$: draw another arc, crossing the previous one at point $Q$.
- Draw the line $PQ$. It is perpendicular to $\ell$, and the intersection with $\ell$ is the foot of the perpendicular.
This is essentially the perpendicular bisector of $XY$, but we use $P$ on one side and $Q$ on the other. $PX = PY$ (radius), $QX = QY$ (radius), so $P$ and $Q$ are both on the perpendicular bisector of $XY$ — and the perpendicular bisector of $XY$ is perpendicular to $\ell$.
Book notes · Perpendicular from a point
- Arc from $P$ crosses line at $X$ and $Y$.
- Equal arcs from $X$ and $Y$ meet at $Q$.
- $PQ$ is the perpendicular.
Watch Me Solve It · 3 examples
- 1Open the compass wideRadius must be more than half of $AB$.
- 2Arcs from $A$Centre on $A$, draw an arc above and below $AB$.
- 3Arcs from $B$ (same radius)Centre on $B$, draw an arc above and below crossing the first arcs.Final step: join the two crossings with the straight-edge. That line is the perpendicular bisector.
- 1Split the angleEach new angle $= \frac{60^{\circ}}{2} = 30^{\circ}$.
- 2Why equal? (1/2)$BP = BQ$ and $PR = QR$ (compass radii).
- 3Why equal? (2/2)$\triangle BPR \equiv \triangle BQR$ (SSS), so $\angle PBR = \angle QBR$.Each is $30^{\circ}$.
- 1Start with a right angleConstruct a perpendicular at a point on a line — this gives a $90^{\circ}$ angle.
- 2Bisect the right angleApply the angle-bisector construction to the $90^{\circ}$ angle.
- 3ResultThe bisector cuts $90^{\circ}$ into two $45^{\circ}$ angles.Two constructions combined: perpendicular + angle bisector.
Common Pitfalls
Perp bisector
- Arcs from $A$, then from $B$ (same radius).
- Join the two crossings.
- Result $\perp AB$ through midpoint.
Angle bisector
- Arc from vertex crosses both arms.
- Equal arcs from those points meet at $R$.
- Ray vertex→$R$ bisects.
Perp from a point
- Arc from $P$ crosses line at $X, Y$.
- Equal arcs from $X, Y$ meet at $Q$.
- $PQ \perp$ line.
Tool rules
- Straight-edge only — no measuring.
- Compass for arcs and circles.
- Leave construction marks.
How are you completing this lesson?
Brain Trainer · 4 problems
Four quick drills on construction methods.
-
1 A perpendicular bisector divides a segment into two pieces of length ____.
Pieces equal each other.Equal halves -
2 An angle bisector of a $80^{\circ}$ angle creates two angles of ____.
$80/2$.$40^{\circ}$ each -
3 Bisecting a right angle yields what angle?
$90/2$.$45^{\circ}$ -
4 Bisecting an angle TWICE divides it into how many equal parts?
$2 \times 2$ pieces.$4$ equal angles
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Describe (in words, no diagram needed) the four steps to construct the perpendicular bisector of a segment $PQ$ using only a straight-edge and a pair of compasses.
Q7. Suppose you've correctly bisected an angle of $130^{\circ}$.
(a) What is the size of each new angle?
(b) If you bisect ONE of those new angles again, what size will THAT angle be?
(c) After two successive bisections, what fraction of the original angle does the final piece represent?
Q8. Explain why the perpendicular-bisector construction works.
(a) Why are the two arcs from $A$ and $B$ drawn with the same radius?
(b) Why does the joining line pass through the midpoint of $AB$?
(c) Why is the joining line perpendicular to $AB$?
Quick Check
1. A — Straight-edge + compasses.
2. D — Cuts $AB$ at $90^{\circ}$ through midpoint.
3. B — More than half $AB$.
4. C — $25^{\circ}$.
5. A — SSS.
Show Your Working Model Answers
Q6 (3 marks): Step 1: Open compass to more than half $PQ$ [1]. Step 2: Centre on $P$, draw an arc above and below the segment. Step 3: Same radius, centre on $Q$, draw arcs above and below crossing the first arcs [1]. Step 4: Use the straight-edge to join the two crossing points; that line is the perpendicular bisector [1].
Q7 (3 marks): (a) $130/2 = 65^{\circ}$ each [1]. (b) $65/2 = 32.5^{\circ}$ [1]. (c) After two bisections each piece is $\frac{1}{4}$ of the original angle [1].
Q8 (3 marks): (a) Equal radii guarantee both intersection points are EQUIDISTANT from $A$ and $B$ [1]. (b) Points equidistant from $A$ and $B$ lie on the perpendicular bisector, which by definition passes through the midpoint of $AB$ [1]. (c) The two equal radii form rhombuses; the diagonals of a rhombus are perpendicular — so the joining line is perpendicular to $AB$ [1].
Construct an Equilateral Triangle
You're given a line segment $AB$ of length $5$ cm. Using only a straight-edge and compasses, you need to construct an equilateral triangle on $AB$ (all sides equal $5$ cm). (a) Describe the construction step-by-step. (b) Use the SSS congruence rule to explain why the triangle MUST be equilateral. (c) How would you then construct a $30^{\circ}$ angle from this figure?
Reveal solution
(a) Set compass radius to length $AB$ ($5$ cm). Centre on $A$ — arc above. Same radius, centre on $B$ — arc above crossing the first arc at $C$. Join $AC$ and $BC$. (b) $AB = AC = BC$ (all equal to the compass radius), so $\triangle ABC$ is equilateral by definition. By SSS its three angles are all equal, and since they sum to $180^{\circ}$, each is $60^{\circ}$. (c) Bisect any of the $60^{\circ}$ angles — the bisector creates two $30^{\circ}$ angles.
Tools
Straight-edge + pair of compasses.
Perp bisector
Same-radius arcs from $A$ and $B$; join crossings.
Angle bisector
Arc from vertex; equal arcs from arms meet at $R$; join.
Perp from point
Like perp bisector applied to two crossing points.
Why it works
Equal radii $\Rightarrow$ congruent triangles (SSS).
Leave arcs
Arcs are evidence of method — don't erase.
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