CONSTRUCTIONS (Using ruler and compasses only)
174. The Angle Splitter · Bisect any angle — and discover the equidistance property
Ray BF always bisects ∠ABC — ∠ABF = ∠FBC.
The bisector of an angle is the ray from the vertex that divides the angle into two equal parts. To bisect ∠ABC: draw an arc centred at B (cutting arms at D and E), then draw equal arcs from D and E to meet at F. Ray BF is the bisector.
Class-8 key property — Equidistance: Every point on the angle bisector is equidistant from both arms (measured as perpendicular distance).
Incenter: The bisectors of all three angles of a triangle are concurrent at the incenter I, which is equidistant from all three sides. Formula: ∠BIC = 90° + ∠A/2.
What this lesson covers
Try to break it
Drag A or C to change ∠ABC. BF stays true: ∠ABF = ∠FBC always. The deeper idea new in class 8 is equidistance — any point on the bisector is equally far (perpendicularly) from both arms. That is why the incenter I of a triangle (where all three bisectors meet) is equidistant from all three sides, and ∠BIC = 90° + ∠A/2.
How you build it
Construct the bisector of an angle using compass and straightedge.
- Draw ray BA — the first arm of ∠ABC from vertex B.
- Draw ray BC — the second arm. Rays BA and BC together form ∠ABC.
- With centre B and any radius r₁, draw an arc cutting arm BA at D and arm BC at E. Both D and E are at equal distance r₁ from B.
- With centre D and radius r₂ (greater than half of DE), draw an arc inside the angle.
- With the same radius r₂ and centre E, draw an arc crossing the one from D. The intersection point F has DF = EF — F is equidistant from both arms.
- Draw ray BF. By SSS, △BDF ≅ △BEF ⟹ ∠ABF = ∠FBC. Ray BF bisects ∠ABC, and every point on BF is equidistant (perpendicularly) from arms BA and BC.
The proof, step by step
Prove that ray BF bisects ∠ABC into two equal angles.
- In △BDF and △BEF: BD = BE (radii of the arc centred at B).
- DF = EF (radii of the equal arcs drawn from D and E).
- BF = BF (common side to both triangles).
- ∴ △BDF ≅ △BEF (SSS Congruence Rule).
- ∴ ∠ABF = ∠FBC (CPCT). Hence ray BF bisects ∠ABC.
- Equidistance: since F lies on the bisector, the perpendicular distance from F to arm BA equals the perpendicular distance from F to arm BC.
Worked example
In triangle ABC, ∠A = 50° and ∠B = 70°. The bisectors of ∠B and ∠C meet at I (the incenter). Find ∠BIC.
∠C = 180° − 50° − 70° = 60°. In △BIC: ∠IBC = 35° and ∠ICB = 30°. So ∠BIC = 180° − 35° − 30° = 115°. Using the formula: ∠BIC = 90° + ∠A/2 = 90° + 25° = 115°.
- 115° — correct
- 105°
- 120°
- 125°