High School Maths · Year 2 · Unit 15 · Lesson 1 of 5

Perpendicular and Angle Bisectors: Cut Any Line or Angle Exactly in Half

About 15 minutes · ruler-and-compasses constructions · the perpendicular bisector of a line, step by step · the bisector of an angle · why each construction works (rhombus, congruent triangles) · bisectors in triangles, with Pythagoras · no calculator · exit quiz

Hook

Two villages, A and B, are 8 km apart. The villages want to dig a well that is exactly the same distance from both of them.

Is there only one place where the well can go?

No: there is a whole line of places. The midpoint of AB works, but so does every point on the line that cuts AB in half at 90°. This line is the perpendicular bisector of AB. In this lesson you construct it exactly with a ruler and compasses, and see why it works.

Every point on the perpendicular bisector is the same distance from A and from B.
Learn

Master the essentials

Arcs of the same radius from A and from B cross at C and D. CA = CB = DA = DB, so ACBD is a rhombus.
VP = VQ and PR = QR, so triangles VPR and VQR are congruent (SSS). The two halves of the angle are equal.

Construction kit

A ruler, a sharp pencil and compasses. In a construction you do not use a protractor.

Leave every arc on show: the arcs are your working

Perpendicular bisector

The line that cuts a line segment AB in half and meets it at 90°.

Every point on it is the same distance from A and from B

Bisector: the steps

(1) Open the compasses to MORE than half of AB. (2) From A, draw arcs above and below AB. (3) Same radius: arcs from B. (4) Join the two crossing points.

AB = 7 cm: a radius of 5 cm works; 3 cm does not

Why it works

Each crossing point is the same distance from A as from B. So ACBD is a rhombus, and the diagonals of a rhombus cut each other in half at 90°.

CA = CB = DA = DB

Angle bisector

The line that cuts an angle into two equal angles.

Bisecting 72° gives two angles of 36°

Angle bisector: the steps

(1) From the vertex V, draw an arc cutting both arms, at P and Q. (2) Same radius: arcs from P and from Q that cross at R. (3) Join V to R.

Keep the compasses fixed for steps 1 and 2 if you like

Why it works

VP = VQ, PR = QR and VR is shared, so triangles VPR and VQR are congruent (SSS). The angles at V are equal.

Congruent triangles have equal angles

Perpendicular from a point

To drop a perpendicular from P to a line: an arc from P cuts the line twice; then bisect the gap between those two points.

It gives the shortest distance from P to the line

Worked examples

See it done

1. AB is a line 9 cm long. List, in order, the steps to construct its perpendicular bisector.

  1. Open the compasses to more than half of AB: more than 4.5 cm, say 5.5 cm.
  2. Put the point on A. Draw an arc above AB and an arc below AB.
  3. Keep the same radius. Put the point on B and draw arcs that cross the first two.
  4. Join the two crossing points with a ruler. Leave all the arcs on show.

Equal arcs from A and from B (radius more than 4.5 cm); join the two crossing points

2. Angle PQR = 64°. Describe how to construct its bisector, and give the size of each new angle.

  1. Put the point on Q. Draw an arc that cuts QP and QR.
  2. From each of those two points, draw an arc (same radius) inside the angle, so that the arcs cross.
  3. Join Q to the crossing point.
  4. The bisector cuts 64° in half: 64° ÷ 2 = 32°.

Each new angle is 32°

3. AB = 10 cm. C lies on the perpendicular bisector of AB, 12 cm from the midpoint M of AB. Find the length CA.

  1. The bisector meets AB at 90° at M, and AM = 10 ÷ 2 = 5 cm.
  2. Triangle AMC has a right angle at M.
  3. Pythagoras: CA² = 5² + 12² = 25 + 144 = 169.
  4. CA = √169 = 13 cm. (CB is also 13 cm: C is on the bisector.)

CA = 13 cm

Common mistake: Changing the compass width between the arcs from A and the arcs from B, or using a radius of half of AB or less (then the arcs touch at one point or do not meet at all). Keep one radius, more than half of AB, for all four arcs, and never rub the arcs out: examiners give marks for them.

PractiseNo calculator

10 questions with instant feedback

Choose the correct answer. Picture each construction, step by step, before you decide. Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.Lina is constructing the perpendicular bisector of AB. She has opened her compasses to more than half of AB and drawn arcs above and below AB from A. What should she do next?

Not quite. The radius must stay the same, so each crossing point is the same distance from A and from B.

Solution

  1. The crossing points must be the same distance from A and from B.
  2. So use the same radius from B.
  3. Draw arcs from B that cross the arcs from A, then join the two crossing points.

Answer: B. Keep the same radius and draw arcs above and below AB from B.

Correct.

Solution

  1. The crossing points must be the same distance from A and from B.
  2. So use the same radius from B.
  3. Draw arcs from B that cross the arcs from A, then join the two crossing points.

Not quite. There are no crossing points yet. They appear when the arcs from B are drawn.

Solution

  1. The crossing points must be the same distance from A and from B.
  2. So use the same radius from B.
  3. Draw arcs from B that cross the arcs from A, then join the two crossing points.

Answer: B. Keep the same radius and draw arcs above and below AB from B.

Not quite. Measuring does not construct the bisector, and it does not give the 90° angle.

Solution

  1. The crossing points must be the same distance from A and from B.
  2. So use the same radius from B.
  3. Draw arcs from B that cross the arcs from A, then join the two crossing points.

Answer: B. Keep the same radius and draw arcs above and below AB from B.

2.AB is 11 cm long. Which of these compass radii can be used to construct the perpendicular bisector of AB?

Correct.

Solution

  1. The radius must be more than half of AB.
  2. Half of 11 cm is 5.5 cm.
  3. Only 6 cm is more than 5.5 cm.

Not quite. This is exactly half of AB. The arcs only touch, at the midpoint, so there is no line to draw.

Solution

  1. The radius must be more than half of AB.
  2. Half of 11 cm is 5.5 cm.
  3. Only 6 cm is more than 5.5 cm.

Answer: A. 6 cm.

Not quite. This is less than half of AB (5.5 cm). The arcs from A and B do not meet.

Solution

  1. The radius must be more than half of AB.
  2. Half of 11 cm is 5.5 cm.
  3. Only 6 cm is more than 5.5 cm.

Answer: A. 6 cm.

Not quite. This halves 5.5 cm again. The radius must be MORE than half of AB.

Solution

  1. The radius must be more than half of AB.
  2. Half of 11 cm is 5.5 cm.
  3. Only 6 cm is more than 5.5 cm.

Answer: A. 6 cm.

3.Angle XYZ = 118°. Toby constructs the bisector of angle XYZ. What angle does the bisector make with YX?

Not quite. This is the whole angle. The bisector cuts it in half.

Solution

  1. A bisector cuts the angle into two equal parts.
  2. 118° ÷ 2 = 59°.

Answer: C. 59°.

Not quite. This is 180° − 118°, the angle on a straight line.

Solution

  1. A bisector cuts the angle into two equal parts.
  2. 118° ÷ 2 = 59°.

Answer: C. 59°.

Correct.

Solution

  1. A bisector cuts the angle into two equal parts.
  2. 118° ÷ 2 = 59°.

Not quite. This halves 62°. Halve the angle itself: 118° ÷ 2.

Solution

  1. A bisector cuts the angle into two equal parts.
  2. 118° ÷ 2 = 59°.

Answer: C. 59°.

Exam standard (Several steps, like real exam questions.)

4.Arcs of the same radius are drawn from A and from B and cross at C and D. Which working shows why CD is the perpendicular bisector of AB?

Correct.

Solution

  1. The arcs from A and from B have the same radius, so CA = CB = DA = DB.
  2. A quadrilateral with four equal sides is a rhombus.
  3. The diagonals of a rhombus bisect each other at right angles.
  4. So CD cuts AB in half at 90°.

Not quite. All four sides are equal, so ACBD is a rhombus. Equal diagonals do not give a right angle.

Solution

  1. The arcs from A and from B have the same radius, so CA = CB = DA = DB.
  2. A quadrilateral with four equal sides is a rhombus.
  3. The diagonals of a rhombus bisect each other at right angles.
  4. So CD cuts AB in half at 90°.

Answer: A. CA = CB = DA = DB, so ACBD is a rhombus. The diagonals of a rhombus cut each other in half at 90°.

Not quite. CD need not equal the radius, so triangle ACD is not equilateral in general.

Solution

  1. The arcs from A and from B have the same radius, so CA = CB = DA = DB.
  2. A quadrilateral with four equal sides is a rhombus.
  3. The diagonals of a rhombus bisect each other at right angles.
  4. So CD cuts AB in half at 90°.

Answer: A. CA = CB = DA = DB, so ACBD is a rhombus. The diagonals of a rhombus cut each other in half at 90°.

Not quite. C is the same distance from A and from B, because both arcs have the same radius.

Solution

  1. The arcs from A and from B have the same radius, so CA = CB = DA = DB.
  2. A quadrilateral with four equal sides is a rhombus.
  3. The diagonals of a rhombus bisect each other at right angles.
  4. So CD cuts AB in half at 90°.

Answer: A. CA = CB = DA = DB, so ACBD is a rhombus. The diagonals of a rhombus cut each other in half at 90°.

5.AB = 16 cm. P lies on the perpendicular bisector of AB, 6 cm from AB. Find the length PA.

Not quite. This uses the whole of AB. The bisector meets AB at its midpoint, 8 cm from A.

Solution

  1. The bisector meets AB at its midpoint M at 90°, and AM = 8 cm.
  2. PA² = 8² + 6² = 64 + 36 = 100.
  3. PA = 10 cm.

Answer: C. 10 cm.

Not quite. This subtracts the squares. PA is the hypotenuse, so add them.

Solution

  1. The bisector meets AB at its midpoint M at 90°, and AM = 8 cm.
  2. PA² = 8² + 6² = 64 + 36 = 100.
  3. PA = 10 cm.

Answer: C. 10 cm.

Correct.

Solution

  1. The bisector meets AB at its midpoint M at 90°, and AM = 8 cm.
  2. PA² = 8² + 6² = 64 + 36 = 100.
  3. PA = 10 cm.

Not quite. This adds the lengths: 8 + 6. Use Pythagoras.

Solution

  1. The bisector meets AB at its midpoint M at 90°, and AM = 8 cm.
  2. PA² = 8² + 6² = 64 + 36 = 100.
  3. PA = 10 cm.

Answer: C. 10 cm.

6.Four students each construct the perpendicular bisector of a line AB. The table shows the length of AB and the compass radius each one uses. Whose arcs will NOT cross?

StudentLength of ABCompass radius
Ana8 cm5 cm
Ben12 cm7 cm
Cleo10 cm4.5 cm
Dev6 cm3.5 cm

Each student draws arcs of this radius from A and from B.

Not quite. 5 cm is more than half of 8 cm, so Ana's arcs cross.

Solution

  1. The arcs cross only if the radius is more than half of AB.
  2. Halves: 4 cm, 6 cm, 5 cm and 3 cm.
  3. Cleo uses 4.5 cm, which is less than 5 cm, so her arcs do not cross.

Answer: B. Cleo.

Correct.

Solution

  1. The arcs cross only if the radius is more than half of AB.
  2. Halves: 4 cm, 6 cm, 5 cm and 3 cm.
  3. Cleo uses 4.5 cm, which is less than 5 cm, so her arcs do not cross.

Not quite. 7 cm is more than half of 12 cm, so Ben's arcs cross.

Solution

  1. The arcs cross only if the radius is more than half of AB.
  2. Halves: 4 cm, 6 cm, 5 cm and 3 cm.
  3. Cleo uses 4.5 cm, which is less than 5 cm, so her arcs do not cross.

Answer: B. Cleo.

Not quite. 3.5 cm is more than half of 6 cm, so Dev's arcs cross.

Solution

  1. The arcs cross only if the radius is more than half of AB.
  2. Halves: 4 cm, 6 cm, 5 cm and 3 cm.
  3. Cleo uses 4.5 cm, which is less than 5 cm, so her arcs do not cross.

Answer: B. Cleo.

7.In triangle ABC, angle A = 48° and angle B = 74°. The bisector of angle C meets AB at D. Find angle ADC.

Not quite. This is angle BDC, the other angle at D.

Solution

  1. Angle C = 180° − 48° − 74° = 58°.
  2. The bisector halves it: angle ACD = 29°.
  3. In triangle ADC: angle ADC = 180° − 48° − 29° = 103°.

Answer: B. 103°.

Correct.

Solution

  1. Angle C = 180° − 48° − 74° = 58°.
  2. The bisector halves it: angle ACD = 29°.
  3. In triangle ADC: angle ADC = 180° − 48° − 29° = 103°.

Not quite. This is angle ACD, half of angle C.

Solution

  1. Angle C = 180° − 48° − 74° = 58°.
  2. The bisector halves it: angle ACD = 29°.
  3. In triangle ADC: angle ADC = 180° − 48° − 29° = 103°.

Answer: B. 103°.

Not quite. This takes only angle A from 180°. Angle ACD must go too.

Solution

  1. Angle C = 180° − 48° − 74° = 58°.
  2. The bisector halves it: angle ACD = 29°.
  3. In triangle ADC: angle ADC = 180° − 48° − 29° = 103°.

Answer: B. 103°.

Challenge (Harder problems that mix ideas or need reasoning.)

8.R lies on the perpendicular bisector of PQ. RP = 15 cm, and R is 9 cm from the line PQ. Find the length PQ.

Not quite. This is half of PQ. The bisector meets PQ at its midpoint, so double it.

Solution

  1. Let M be the midpoint of PQ. Triangle RMP has a right angle at M.
  2. PM² = 15² − 9² = 225 − 81 = 144, so PM = 12 cm.
  3. PQ = 2 × 12 = 24 cm.

Answer: D. 24 cm.

Not quite. This adds the squares. RP is the hypotenuse, so subtract.

Solution

  1. Let M be the midpoint of PQ. Triangle RMP has a right angle at M.
  2. PM² = 15² − 9² = 225 − 81 = 144, so PM = 12 cm.
  3. PQ = 2 × 12 = 24 cm.

Answer: D. 24 cm.

Not quite. This doubles RP. Double the half-length found by Pythagoras.

Solution

  1. Let M be the midpoint of PQ. Triangle RMP has a right angle at M.
  2. PM² = 15² − 9² = 225 − 81 = 144, so PM = 12 cm.
  3. PQ = 2 × 12 = 24 cm.

Answer: D. 24 cm.

Correct.

Solution

  1. Let M be the midpoint of PQ. Triangle RMP has a right angle at M.
  2. PM² = 15² − 9² = 225 − 81 = 144, so PM = 12 cm.
  3. PQ = 2 × 12 = 24 cm.

9.In triangle ABC, angle A = 70°. The bisectors of angles B and C meet at I. Find angle BIC.

Not quite. This is angle B + angle C. Each bisector takes only half.

Solution

  1. Angle B + angle C = 180° − 70° = 110°.
  2. Half of each: angle IBC + angle ICB = 110° ÷ 2 = 55°.
  3. Angle BIC = 180° − 55° = 125°.

Answer: D. 125°.

Not quite. This doubles angle A.

Solution

  1. Angle B + angle C = 180° − 70° = 110°.
  2. Half of each: angle IBC + angle ICB = 110° ÷ 2 = 55°.
  3. Angle BIC = 180° − 55° = 125°.

Answer: D. 125°.

Not quite. This is half of angle B + angle C, the angles at B and C in triangle BIC.

Solution

  1. Angle B + angle C = 180° − 70° = 110°.
  2. Half of each: angle IBC + angle ICB = 110° ÷ 2 = 55°.
  3. Angle BIC = 180° − 55° = 125°.

Answer: D. 125°.

Correct.

Solution

  1. Angle B + angle C = 180° − 70° = 110°.
  2. Half of each: angle IBC + angle ICB = 110° ÷ 2 = 55°.
  3. Angle BIC = 180° − 55° = 125°.

10.Mo bisects an angle. Then he bisects one of the two halves. The smallest angle now in his drawing is 17°. What was the size of the first angle?

Correct.

Solution

  1. The second bisection gave 17°, so the half-angle was 2 × 17° = 34°.
  2. The first bisection gave 34°, so the angle was 2 × 34° = 68°.
  3. Check: 68° ÷ 2 = 34° and 34° ÷ 2 = 17° ✓.

Not quite. This undoes only one bisection. Double twice.

Solution

  1. The second bisection gave 17°, so the half-angle was 2 × 17° = 34°.
  2. The first bisection gave 34°, so the angle was 2 × 34° = 68°.
  3. Check: 68° ÷ 2 = 34° and 34° ÷ 2 = 17° ✓.

Answer: A. 68°.

Not quite. This halves 17°. Work backwards: double it.

Solution

  1. The second bisection gave 17°, so the half-angle was 2 × 17° = 34°.
  2. The first bisection gave 34°, so the angle was 2 × 34° = 68°.
  3. Check: 68° ÷ 2 = 34° and 34° ÷ 2 = 17° ✓.

Answer: A. 68°.

Not quite. This counts three equal parts. Two bisections give four quarter-angles.

Solution

  1. The second bisection gave 17°, so the half-angle was 2 × 17° = 34°.
  2. The first bisection gave 34°, so the angle was 2 × 34° = 68°.
  3. Check: 68° ÷ 2 = 34° and 34° ÷ 2 = 17° ✓.

Answer: A. 68°.

Apply

Construct it, then check it

Use a ruler, a pencil and compasses. Answer each question on paper, then compare with the model.

1. Draw a line PQ 7.4 cm long. Construct its perpendicular bisector, leaving your arcs. Measure to check that it cuts PQ at its midpoint.

2. Draw an angle of any size. Construct its bisector. Check both halves with a protractor.

3. Explain why every point on the perpendicular bisector of PQ is the same distance from P and from Q.

  1. Open the compasses to more than 3.7 cm, say 5 cm.
  2. Draw arcs above and below PQ from P, then from Q with the same radius.
  3. Join the two crossing points. It should cut PQ at 3.7 cm from each end, at 90° (within 2 mm and 2°).
  1. Arc from the vertex cutting both arms.
  2. Equal arcs from those two points, crossing inside the angle.
  3. Join the vertex to the crossing point. Both halves should measure the same.
  1. Take any point X on the bisector, and let M be the midpoint of PQ.
  2. Triangles XMP and XMQ have MP = MQ, a shared side XM and right angles at M.
  3. So they are congruent (SAS), and XP = XQ ✓.

Next: the Lesson 1 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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