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BISSECTING ANGLES

Meaning / Definition

To bisect means to divide into two equal parts.

So bisecting an angle means drawing a line (or ray) that splits the given angle into two equal angles.

The line or ray that does the division is called the angle bisector or bisector of that angle.

After bisecting, each of the two smaller angles has the same measure.

For example, if you have an angle of 60°, and you bisect it, each part becomes 30°.



Why Bisect Angles?
  1. It helps us to divide angles into exact halves without measuring with a protractor.

  2. It is used in construction problems and many geometry proofs.

  3. It is used when we want special angles like 15°, 30°, 45° (by bisecting 30°, 60°, 90°, etc.).

  4. In triangles, angle bisectors have special properties (e.g. angle bisector theorem).

Tools Needed

To do angle bisection you need:

  • A compass

  • A ruler (straightedge)

  • Pencil

  • (Optional: protractor, for verification, but not needed in the construction)


Properties of the Angle Bisector
  • The bisector divides the angle into two equal angles.

  • Any point on the bisector is equidistant from the two sides (arms) of the angle.

  • The angle bisector in a triangle divides the opposite side in the ratio of the other two sides (Angle Bisector Theorem).

  • You can bisect any angle — acute, obtuse, or right angle.


Step-by-Step Construction: Bisecting a Given Angle

Suppose we have an angle ∠ABC (vertex at B, rays BA and BC).

Goal: Construct the bisector of ∠ABC, that is a ray from B that divides ∠ABC into two equal angles.

Steps:

  1. Place the compass point at B (the vertex). Open it to some convenient radius (not too small, not too big).

  2. With center B, draw an arc that cuts both arms BA and BC. Let the arc cut BA at point D and BC at point E.

  3. Without changing the compass width, place the compass at D and draw an arc inside the angle.

  4. Keeping the same compass width, place the compass at E and draw another arc inside the angle, so that this arc intersects the one from D. Label their intersection F.

  5. Use your ruler to draw ray BF. That is the angle bisector of ∠ABC.

Text Diagram (ASCII-style):

       C
      /
     / 
    E
   /  
  /   F
 B----D
  \
   \
    A

B is the vertex.

The arcs from D and E meet at F.

BF is the bisector.

After construction, ∠DBF = ∠FBC (they are equal).

Explanation / Why It Works (Proof Sketch)

When you draw the first arc from B that meets the sides at D and E, BD = BE (both are arcs from B).

Then you draw arcs from D and E of the same radius, so the intersection point F is equidistant from D and E.

In triangles BDF and BEF, we have:

  • BD = BE
  • DF = EF
  • BF is common

So by SSS (side-side-side), triangles BDF and BEF are congruent. Hence ∠DBF = ∠FBE, so BF bisects ∠DBE (i.e., original angle).





Practice Questions

Problem 1 — Bisect an acute angle (70°)

Question: Bisect ∠XYZ (given as 70° on the diagram).

Steps

  1. Put the compass point at the vertex Y. Open it to a convenient width.

  2. With centre Y, draw an arc that cuts YX at A and YZ at B.

  3. Without changing the compass width, put the point at A and draw an arc inside the angle.

  4. With the same compass width put the point at B and draw another arc so it meets the arc from A at C.

  5. Use the ruler to join Y to C. Ray YC is the bisector.

Diagram

    Z
     \
      \
       B
        \
         C
        /
 Y----A/
      /
     X

Check: Measure ∠XYC and ∠CYZ with a protractor — both should be 35° (half of 70°).

Answer: YC bisects ∠XYZ; ∠XYC = ∠CYZ = 35°.

Problem 2 — Bisect a right angle (90°)

Question: Bisect the right angle at vertex P to get 45°.

Steps

  1. Place the compass at P. Draw an arc that meets both arms at D and E.

  2. With the same compass width, draw an arc from D inside the angle.

  3. With the same width draw an arc from E so the two arcs meet at F.

  4. Join P to F. Ray PF bisects the 90° angle.

Diagram

    |
    |
 F  |
    |
P---D---->  (other arm)
    \
     E

Check: ∠(one arm, PF) should be 45°; verify with protractor.

Answer: PF bisects the 90° angle; each part = 45°.

Problem 3 — Bisect an obtuse angle (120°)

Question: Bisect ∠AOB that is 120°.

Steps

  1. Put compass at O and draw an arc that cuts OA at U and OB at V.

  2. With the same compass width draw an arc from U inside the angle.

  3. With the same width draw an arc from V so the two arcs meet at W.

  4. Join O to W. Ray OW bisects ∠AOB.

Diagram

 A
  \
   \
    U
     \
      \
       W
      /
     /
 O--V
    \
     \
      B

Check: Each smaller angle should be 60° (since 120° ÷ 2 = 60°). Measure to confirm.

Answer: OW bisects the obtuse angle into two 60° angles.

Problem 4 — Make 30° by bisecting 60°

Question: Construct a 60° angle and then bisect it to get a 30° angle.

Steps to make 60°

  1. Draw a ray OP (baseline). Mark a point Q on OP.

  2. With centre O and radius OQ draw an arc that meets OP at Q.

  3. With same radius and centre Q draw an arc that meets the arc from O at R.

  4. Join O to R — ∠QOR = 60°.

Steps to bisect 60°

  1. With centre O draw an arc that cuts OR at S and OQ at T.

  2. With same radius draw arc from S, and from T, their intersection is U.

  3. Join O to U. ∠QOU = 30°.

Diagram

     R
    /
   /
  O---Q----------
   \
    \
     U

Check: Measure ∠QOU — it should be 30°.

Answer: Constructed 60° then bisected to get 30°.

Problem 5 — Bisect an angle in a triangle (bisector from vertex A)

Question: In triangle ABC, construct the bisector of angle A and let it meet BC at D.

Steps

  1. At vertex A, put the compass and draw an arc cutting AB at E and AC at F.

  2. With the same compass width draw an arc from E inside the triangle.

  3. With the same width draw an arc from F so the arcs meet at G.

  4. Join A to G; this line meets BC at D. AD is the angle bisector of ∠A.

Diagram

    C
   / \
  /   \
 A-----\
 | \    \
 |  \    \
 E   G    F
 |    \   |
 B-----D--

Check: Measure ∠BAD and ∠DAC; they should be equal. Also AD meets BC at D.

Answer: AD bisects ∠A and meets BC at D.

Problem 6 — Construct three angle bisectors of a triangle (find the in-center)

Question: For triangle PQR, construct the bisectors of ∠P, ∠Q and ∠R. Show that the three bisectors meet at one point (the in-center).

Steps

  1. Bisect ∠P using the standard arc method to get ray p.

  2. Bisect ∠Q to get ray q.

  3. The intersection of rays p and q is point I.

  4. (Optional) Bisect ∠R — its bisector should also pass through I.

  5. Point I is the in-center (the point where all three internal bisectors meet).

Diagram

     R
    / \
   /   \
  /     \
 P---I---Q

Check: After constructing I, draw the bisector of R; it should pass through I. You can also draw a small circle centred at I that touches each side (this is the incircle) — check by measuring distance from I to each side (equal).

Answer: The three internal bisectors meet at I (the in-center).

Problem 7 — Construct 15° by repeated bisection

Question: Construct a 15° angle (use only ruler and compass).

Steps

  1. First construct a 60° angle (use the equilateral/arc method as in Problem 4).

  2. Bisect 60° to get 30° (use the angle bisection steps).

  3. Bisect the 30° angle: draw arcs from the points where the 30° arc cuts its sides, find intersection and join to the vertex — the new ray makes 15° with the baseline.

Diagram

    (top)
     R
    /
   /
  O---Q---------
   \
    \
     U  (gives 15°)

Check: Use a protractor to confirm the final angle is 15°. Or check by noting 60° → 30° → 15° by successive halving.

Answer: Final ray gives angle of 15° with the base.




CHECK OTHER RELATED TOPICS HERE


  1. ANGLES

  2. BEARING


  3. CONSTRUCTION

  4. BISECTING ANGLES


  5. DATA PRESENTATION
    6.
  6. PROBABILITY




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