Construction: Bisector of an Angle
Constructing the bisector of an angle is one of the fundamental geometric constructions using a compass and straightedge (ruler without markings). An angle bisector is a ray that divides an angle into two equal parts.
In NCERT Class 9 Mathematics (Chapter 11 — Constructions), students learn to construct angle bisectors with proper justification. Unlike measurement-based constructions, these constructions use only a compass and a straightedge — no protractor is used.
The ability to bisect an angle is essential for constructing specific angles (such as 30°, 45°, 15°), bisecting triangles, and creating symmetric designs. The construction is also the basis for constructing perpendicular bisectors and other advanced figures.
What is Construction: Bisector of an Angle?
Definition: The bisector of an angle is a ray that originates from the vertex of the angle and divides it into two equal angles.
If ∠AOB = θ, then the bisector OC divides it into:
∠AOC = ∠COB = θ/2
Key properties of the angle bisector:
- Every point on the angle bisector is equidistant from the two arms (sides) of the angle.
- The bisector of an angle is unique.
- The three angle bisectors of a triangle meet at a single point called the incentre.
- The incentre is equidistant from all three sides (it is the centre of the incircle).
Tools required:
- A compass (for drawing arcs)
- A straightedge (ruler without markings, for drawing lines)
- A pencil
Construction: Bisector of an Angle Formula
Construction Steps — Bisecting a Given Angle:
Given: An angle ∠AOB with vertex O.
To construct: The bisector of ∠AOB.
Steps:
- With O as centre and any convenient radius, draw an arc that intersects ray OA at point P and ray OB at point Q.
- With P as centre and a radius greater than half of PQ, draw an arc in the interior of the angle.
- With Q as centre and the same radius, draw another arc intersecting the previous arc at point R.
- Draw ray OR.
Ray OR is the bisector of ∠AOB.
Result: ∠AOR = ∠BOR = ½ ∠AOB.
Derivation and Proof
Justification (Proof) of the Construction:
To prove: ∠AOR = ∠BOR (i.e., OR bisects ∠AOB).
Proof:
- In the construction, OP = OQ (radii of the same arc from step 1).
- PR = QR (arcs in steps 2 and 3 were drawn with the same radius from P and Q).
- OR = OR (common side).
- Therefore, ▵OPR ≅ ▵OQR (by SSS congruence).
- By CPCT: ∠POR = ∠QOR.
- That is, ∠AOR = ∠BOR.
Hence, ray OR is the bisector of ∠AOB. ◻
Why does this work?
- Points P and Q are equidistant from O (same arc radius).
- Point R is equidistant from P and Q (same compass opening).
- Therefore, R lies on the perpendicular bisector of PQ, which passes through O.
- The line OR makes equal angles with OP and OQ by the congruence proved above.
Types and Properties
Applications and extensions of angle bisection:
1. Constructing a 30° angle
- First construct a 60° angle (equilateral triangle method).
- Then bisect 60° to get 30°.
2. Constructing a 45° angle
- First construct a 90° angle (perpendicular).
- Then bisect 90° to get 45°.
3. Constructing a 15° angle
- Construct 60°, bisect to get 30°, bisect again to get 15°.
4. Constructing a 75° angle
- Construct 60° and 90° adjacent to each other.
- Bisect the 30° angle between them to get 75° from one of the arms.
5. Angle bisector of a triangle
- The angle bisectors of a triangle are concurrent at the incentre.
- The incentre is equidistant from all three sides.
- A circle drawn from the incentre tangent to all three sides is the incircle.
6. Repeated bisection
- Any angle can be repeatedly bisected to get smaller angles.
- 60° → 30° → 15° → 7.5° → ...
Solved Examples
Example 1: Example 1: Bisecting a 60° angle
Problem: Construct a 60° angle and then bisect it to get a 30° angle.
Solution:
Step 1: Construct 60°
- Draw a ray OA.
- With O as centre and any radius, draw an arc cutting OA at P.
- With P as centre and the same radius, draw an arc cutting the first arc at Q.
- Join OQ and extend. ∠AOQ = 60°.
Step 2: Bisect 60°
- With P as centre and radius > ½ PQ, draw an arc.
- With Q as centre and the same radius, draw an arc intersecting at R.
- Draw ray OR.
Result: ∠AOR = 30°.
Example 2: Example 2: Bisecting a 90° angle
Problem: Construct a 90° angle and bisect it to get a 45° angle.
Solution:
Step 1: Construct 90°
- Draw a ray OA.
- Construct a 60° angle (as in Example 1), getting point Q on the arc.
- With Q as centre and the same radius, draw an arc cutting the first arc at S.
- Bisect the arc QS — this gives a point on the 90° ray. Draw ray OB. ∠AOB = 90°.
Step 2: Bisect 90°
- Apply the standard angle bisection procedure to ∠AOB.
Result: Each half = 45°.
Example 3: Example 3: Bisecting an angle of 120°
Problem: Construct a 120° angle and bisect it.
Solution:
Step 1: Construct 120°
- Draw a ray OA.
- With O as centre, draw an arc cutting OA at P.
- With P as centre and the same radius, mark Q on the arc (60°).
- With Q as centre and the same radius, mark S on the arc (120°).
- Join OS. ∠AOS = 120°.
Step 2: Bisect 120°
- P and S are on the arc. With P and S as centres, draw arcs with equal radii intersecting at R.
- Draw ray OR.
Result: ∠AOR = ∠ROS = 60°.
Example 4: Example 4: Constructing 15° angle
Problem: Construct a 15° angle using compass and straightedge only.
Solution:
- Construct 60° using the equilateral triangle method.
- Bisect 60° to get 30°.
- Bisect 30° to get 15°.
Result: The angle of 15° is obtained by two successive bisections of 60°.
Example 5: Example 5: Bisecting an obtuse angle
Problem: An angle ∠XOY measures 140°. Describe how to bisect it.
Solution:
The procedure is identical for any angle (acute, right, or obtuse):
- With O as centre, draw an arc cutting OX at P and OY at Q.
- With P and Q as centres and equal radii (> ½PQ), draw arcs intersecting at R inside the angle.
- Draw ray OR.
Result: ∠XOR = ∠YOR = 70°.
Note: The same method works for angles of any measure between 0° and 360°.
Example 6: Example 6: Constructing 75° angle
Problem: Construct a 75° angle using compass and straightedge.
Solution:
- Draw a ray OA.
- Construct 60°: Mark point B on the arc such that ∠AOB = 60°.
- Construct 90°: Use the 60° and bisection methods to mark point C such that ∠AOC = 90°.
- Bisect ∠BOC (which is 30°):
- The bisector gives a ray at 60° + 15° = 75° from OA.
Result: ∠ = 75°.
Example 7: Example 7: Triangle angle bisectors meeting at incentre
Problem: In ▵ABC with ∠A = 60°, ∠B = 80°, ∠C = 40°, find the angle between the bisectors of ∠B and ∠C.
Solution:
Given:
- ∠B = 80° ⇒ half = 40°
- ∠C = 40° ⇒ half = 20°
In ▵BIC (where I is the incentre):
- ∠IBC = 40°, ∠ICB = 20°
- ∠BIC = 180° − 40° − 20° = 120°
General formula: ∠BIC = 90° + ½ ∠A
- = 90° + 30° = 120° ✓
Answer: The angle between the bisectors = 120°.
Example 8: Example 8: Bisecting a straight angle
Problem: Bisect a straight angle (180°) to construct a 90° angle (perpendicular).
Solution:
- Draw a straight line AB through point O.
- With O as centre, draw an arc of convenient radius intersecting the line at P (on OA) and Q (on OB).
- With P and Q as centres and a radius greater than half of PQ, draw arcs above the line intersecting at R.
- Draw ray OR.
Result: ∠AOR = ∠BOR = 90°. OR is perpendicular to AB.
Example 9: Example 9: Error in construction
Problem: A student constructed the bisector of an angle but used different radii from P and Q in step 3. Is the bisector correct?
Solution:
Analysis:
- The justification requires ▵OPR ≅ ▵OQR by SSS.
- For SSS: OP = OQ (same arc), OR = OR (common), and PR must equal QR.
- If different radii are used from P and Q, then PR ≠ QR.
- The SSS condition fails, and the line OR does not bisect the angle.
Answer: No. The radii from P and Q must be equal for a correct bisection.
Example 10: Example 10: Constructing the incircle of a triangle
Problem: Describe the steps to construct the incircle of ▵ABC.
Solution:
- Bisect ∠A: Using the standard angle bisection method, draw the bisector of ∠A.
- Bisect ∠B: Similarly, draw the bisector of ∠B.
- The two bisectors intersect at point I (the incentre).
- Drop a perpendicular from I to any side (say AB) to find the point of tangency M. The length IM is the inradius.
- With I as centre and IM as radius, draw the incircle.
Result: The incircle touches all three sides of the triangle.
Real-World Applications
Applications of Angle Bisector Construction:
- Constructing standard angles: Bisecting 60° gives 30°; bisecting 90° gives 45°; bisecting 30° gives 15°. These are used throughout geometry.
- Constructing the incentre and incircle: The bisectors of all three angles of a triangle meet at the incentre, which is the centre of the largest circle fitting inside the triangle.
- Symmetry in design: Angle bisection creates symmetric shapes used in architectural patterns, logos, and decorative arts.
- Dividing land or paths: Bisecting an angle at a junction divides the region into two equal angular sections.
- Engineering: Gear teeth and mechanical components require precise angle division.
- Paper folding (origami): Folding a corner to match the opposite edge creates an angle bisector.
Key Points to Remember
- An angle bisector divides an angle into two equal parts.
- Construction requires only a compass and straightedge — no protractor.
- Steps: Draw an arc from the vertex; from the two intersection points, draw arcs with equal radii; the intersection gives a point on the bisector.
- The construction is justified using SSS congruence of two triangles.
- The radii used from points P and Q must be equal.
- Every point on the bisector is equidistant from the two arms of the angle.
- The three angle bisectors of a triangle meet at the incentre.
- Bisecting 60° gives 30°; bisecting 90° gives 45°.
- The same construction works for any angle — acute, right, obtuse, or straight.
- This is one of the basic constructions required in the CBSE Class 9 syllabus.
Practice Problems
- Construct a 60° angle and bisect it. Verify that each part measures 30°.
- Construct a 90° angle using the perpendicular method, then bisect it to get 45°.
- Construct a 135° angle and bisect it. What angles do you get?
- Construct angles of 15° and 75° using only compass and straightedge.
- Draw any acute angle. Bisect it. Then bisect one of the halves. How many parts have you created?
- Construct ▵ABC with ∠A = 60°, AB = 6 cm, AC = 5 cm. Draw the bisector of ∠A.
- In ▵PQR with ∠P = 50°, ∠Q = 70°, ∠R = 60°, find the angle between the bisectors of ∠Q and ∠R at the incentre.
- Explain why a protractor cannot replace the compass-straightedge method in formal constructions.
Frequently Asked Questions
Q1. What is an angle bisector?
An angle bisector is a ray that divides an angle into two equal parts. It originates from the vertex of the angle.
Q2. What tools are needed to bisect an angle?
Only a compass and a straightedge (ruler without markings) are needed. No protractor is used in formal geometric constructions.
Q3. How do you bisect an angle using a compass?
Draw an arc from the vertex cutting both arms. From the two intersection points, draw arcs with equal radii. The arcs intersect at a point. Join this point to the vertex — that ray is the bisector.
Q4. Why must the radii from P and Q be equal?
Equal radii ensure PR = QR, which is needed for the SSS congruence proof. Without equal radii, the two triangles are not congruent, and the line does not bisect the angle.
Q5. Can you bisect any angle?
Yes. The construction works for any angle — acute, right, obtuse, straight, or reflex. The procedure is the same regardless of the angle size.
Q6. What is the incentre of a triangle?
The incentre is the point where the three angle bisectors of a triangle meet. It is equidistant from all three sides and is the centre of the incircle (the largest circle that fits inside the triangle).
Q7. How do you construct a 45° angle?
Construct a 90° angle (perpendicular), then bisect it. Each half measures 45°.
Q8. What is the justification for the angle bisector construction?
The justification uses SSS congruence. The two triangles formed by the vertex, the arc points, and the intersection point are congruent (equal sides), proving that the angles formed are equal.
Q9. Is angle bisector construction in CBSE Class 9?
Yes. Constructing the bisector of a given angle is part of Chapter 11 (Constructions) in the CBSE Class 9 syllabus. Students must know the steps and the justification.
Q10. What is the formula for the angle at the incentre?
The angle formed between the bisectors of angles B and C at the incentre I is: ∠BIC = 90° + ½∠A. This is a useful result for triangle problems.
