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Lesson 19 ~25 min Unit 3 · Geometry +85 XP

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.

Today's hook: No protractor allowed! Can you cut an angle exactly in half using nothing but a pair of compasses and a ruler? Yes — and it's beautiful.
0/5QUESTS
Think First
warm-up

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?

Record your answer in your workbook.
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What Is a Construction?
+5 XP

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:

  1. Perpendicular bisector of a line segment (cuts the segment in half AT right angles).
  2. Angle bisector (cuts an angle exactly in half).
  3. 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!).

Tools: straight-edge + compasses Ruler (straight-edge) Compass arc No protractor — no marked measurements
Constructions: exact geometry with just two tools
Use the EDGE
A ruler is just a straight-edge in constructions — never measure with it.
Keep the radius
For many constructions you mustn't change the compass opening between arcs.
Leave arcs in
Show your construction marks (arcs) — they're evidence of method.
2
What You'll Master
objectives

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
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Words You Need
vocabulary
ConstructionDrawing exact shapes with only straight-edge + compasses.
Straight-edgeA ruler used only for its edge, not its scale.
Compass / CompassesThe tool for drawing circles — "pair of compasses".
ArcPart of a circle drawn by the compass.
Perpendicular bisectorA line that cuts a segment in half at a right angle.
Angle bisectorA ray that cuts an angle into two equal parts.
MidpointThe exact middle of a segment.
PerpendicularAt a right angle ($90^{\circ}$).
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Construct the Perpendicular Bisector
+5 XP

Given a segment $AB$. To find a line that cuts $AB$ exactly in half at a right angle:

  1. Place the compass point at $A$. Open it to MORE than half the length of $AB$.
  2. Draw a large arc that goes above AND below the segment.
  3. Without changing the radius, move the compass point to $B$.
  4. Draw a second arc that crosses the first arc above AND below.
  5. 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.

Perpendicular bisector of AB A B midpoint perpendicular bisector
Same radius from $A$ and $B$ ⇒ meets on perp bisector
Open WIDE
Radius must be MORE than half $AB$ or the arcs won't meet.
Same radius
Don't change the compass when swapping from $A$ to $B$.
Two intersections
Join the two crossings above and below the segment.
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.
In a perpendicular-bisector construction, the compass radius should be:
5
Construct the Angle Bisector
+5 XP

Given an angle $\angle ABC$ with vertex at $B$. To draw a ray from $B$ that cuts $\angle ABC$ exactly in half:

  1. Place the compass point at $B$. Choose any radius.
  2. Draw an arc that crosses BOTH sides of the angle. Label the crossings $P$ (on $BA$) and $Q$ (on $BC$).
  3. Without changing the radius, place the compass at $P$ and draw an arc inside the angle.
  4. Same radius from $Q$; draw another arc inside, crossing the previous one at a point $R$.
  5. 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.

Angle bisector of ∠ABC B A C Q P R bisector
$\angle ABR = \angle RBC$  (SSS ⇒ equal angles)
Vertex first
Always start with compass on the vertex.
One radius for $P$/$Q$, another for $R$
You can change radius between step 2 and step 3 — just keep it the same in step 3 and 4.
Draw to $R$
The bisector ray goes from the vertex through the intersection $R$.
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.
True or false?

The angle bisector divides an angle into two angles that are equal in measure.

6
Perpendicular From a Point
+5 XP

Given a line $\ell$ and a point $P$ NOT on $\ell$. To drop a perpendicular from $P$ onto $\ell$:

  1. Compass on $P$. Open wide enough so an arc crosses $\ell$ at two points; call them $X$ and $Y$.
  2. Same (or larger) radius from $X$: draw an arc on the opposite side of $\ell$ from $P$.
  3. Same radius from $Y$: draw another arc, crossing the previous one at point $Q$.
  4. 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 $\ell$.

Drop a perpendicular from P P X Y Q
$PQ \perp \ell$ at the foot
Two crossings of $\ell$
The first arc MUST cross $\ell$ twice — open wider if not.
Other side of line
The arcs from $X$ and $Y$ should cross on the OPPOSITE side of $\ell$ from $P$.
Same radius
Steps 2 and 3 must use the same radius.
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.
The two tools used in a classical construction are a compass and a __________.
Watch Me Solve It · Steps for perpendicular bisector
+15 XP per step
Q1
PROBLEM
List the FOUR construction steps to draw the perpendicular bisector of a segment $AB$.
  1. 1
    Open the compass wide
    Radius must be more than half of $AB$.
  2. 2
    Arcs from $A$
    Centre on $A$, draw an arc above and below $AB$.
  3. 3
    Arcs 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.
AnswerSet radius, arc from $A$, arc from $B$, join crossings.
Watch Me Solve It · Angle bisector for $60^{\circ}$
+15 XP per step
Q2
PROBLEM
You construct the bisector of a $60^{\circ}$ angle. What is the size of each of the two new angles, and why are they equal?
  1. 1
    Split the angle
    Each new angle $= \frac{60^{\circ}}{2} = 30^{\circ}$.
  2. 2
    Why equal? (1/2)
    $BP = BQ$ and $PR = QR$ (compass radii).
  3. 3
    Why equal? (2/2)
    $\triangle BPR \equiv \triangle BQR$ (SSS), so $\angle PBR = \angle QBR$.
    Each is $30^{\circ}$.
AnswerEach half is $30^{\circ}$ — equal by SSS congruence.
Watch Me Solve It · Construct a $45^{\circ}$ angle
+15 XP per step
Q3
PROBLEM
Without a protractor, how can you construct a $45^{\circ}$ angle using only the constructions from this lesson?
  1. 1
    Start with a right angle
    Construct a perpendicular at a point on a line — this gives a $90^{\circ}$ angle.
  2. 2
    Bisect the right angle
    Apply the angle-bisector construction to the $90^{\circ}$ angle.
  3. 3
    Result
    The bisector cuts $90^{\circ}$ into two $45^{\circ}$ angles.
    Two constructions combined: perpendicular + angle bisector.
AnswerConstruct $90^{\circ}$, then bisect → two $45^{\circ}$ angles.
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Common Pitfalls
heads-up
Changing the compass radius too early
If you change the radius between the two arcs (from $A$ and from $B$) in a perpendicular bisector, the construction fails.
Fix: Set the radius once; don't touch the compass until BOTH arcs are drawn.
Radius too small
If the radius is less than half $AB$, the arcs from $A$ and $B$ won't intersect.
Fix: Aim for about $\frac{3}{4}$ of the length of $AB$ — that's safely "more than half".
Erasing the arcs
In exams, "show your construction" means leave the arcs! They are evidence of the method.
Fix: Draw lightly so arcs are visible but not messy — never erase them.
Copy Into Your Books

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?

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Brain Trainer · Constructions
4 problems

Four quick drills on construction methods.

  1. 1 A perpendicular bisector divides a segment into two pieces of length ____.

    Pieces equal each other.Equal halves
  2. 2 An angle bisector of a $80^{\circ}$ angle creates two angles of ____.

    $80/2$.$40^{\circ}$ each
  3. 3 Bisecting a right angle yields what angle?

    $90/2$.$45^{\circ}$
  4. 4 Bisecting an angle TWICE divides it into how many equal parts?

    $2 \times 2$ pieces.$4$ equal angles
Complete in your workbook.
1
A classical geometric construction uses ONLY:
+10 XP
2
A perpendicular bisector of $AB$ is a line that:
+10 XP
3
For the perpendicular bisector construction, the compass opening must be:
+10 XP
4
The angle bisector of a $50^{\circ}$ angle creates two angles each of:
+10 XP
5
In the angle-bisector construction, the two triangles formed are congruent by:
+10 XP
Show Your Working
9 marks total
Recall Easy 3 MARKS

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.

Answer in your workbook.
Apply Medium 3 MARKS

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?

Answer in your workbook.
Reason Hard 3 MARKS

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 $AB$?

Answer in your workbook.
Comprehensive Answers

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].

Stretch Challenge · +25 XP, +10 coins

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.

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Quick Review

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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