Constructions

134. Splitting the Angle · The art of making two equal halves

The bisector of an angle is the ray from the vertex that divides the angle into two equal halves. Here, ray BF bisects ∠ABC, so ∠ABF = ∠FBC.

The bisector of an angle is a ray drawn from the vertex that divides the angle into two equal halves. If ray BF bisects ∠ABC, then ∠ABF = ∠FBC, and each is exactly half of ∠ABC. Every angle has one and only one bisector, and it always lies inside the angle. A useful property worth remembering: every point on the angle bisector is equidistant from the two arms of the angle.

Stuck? Ask Guru

Selina ICSE: Constructions

What this lesson covers

Try to break it

Drag A and C to change the angle. Why does ray BF always cut it into two equal halves? Look at triangles BDF and BEF: BD = BE (both are radii of the arc centred at B), DF = EF (both come from the equal arcs drawn from D and E), and BF is common to both. So △BDF ≅ △BEF by SSS — and therefore ∠ABF = ∠FBC. The construction can't fail.

How you build it

Construct the bisector of an angle.

Mark point B — the vertex of the angle you want to bisect.
Draw a ray from B — the first arm of the angle (call its direction BA).
Draw a second ray from B in a different direction — the other arm (BC). The two rays together form ∠ABC, the angle we will bisect.
With centre B and any convenient radius, draw an arc that cuts BOTH arms of the angle.
Mark point D where the arc cuts the first arm (BA). Note: BD equals the arc''s radius.
Mark point E where the arc cuts the other arm (BC). Because both D and E lie on the same arc centred at B, BD = BE.
With the same compass opening and centre D, draw an arc inside the angle.
With the same opening and centre E, draw another arc that crosses the one from D. Because both arcs share the same radius, the meeting point is equidistant from D and E.
Mark point F where the two arcs cross. Because both arcs had the same radius, DF = EF.
Draw the ray BF. In triangles BDF and BEF: BD = BE, DF = EF, and BF is common — so △BDF ≅ △BEF by SSS. Therefore ∠ABF = ∠FBC, which means BF bisects ∠ABC.

The proof, step by step

Prove that ray BF bisects angle ABC into two equal angles.

In △BDF and △BEF: BD = BE (radii of the arc drawn with centre B).
DF = EF (radii of the equal arcs drawn from D and E).
BF = BF (common to both triangles).
∴ △BDF ≅ △BEF (by SSS Congruence Rule).
∴ ∠ABF = ∠FBC (CPCT).
Hence, ray BF bisects ∠ABC.

Worked example

In triangle ABC, angle B = 60° and angle C = 80°. The bisectors of angle B and angle C intersect at O. Find the measure of angle BOC.

In triangle ABC, angle A = 180° - (60° + 80°) = 40°. In triangle BOC, angle OBC = 30° and angle OCB = 40°. So angle BOC = 180° - (30° + 40°) = 110°. Alternatively, angle BOC = 90° + angle A/2 = 90° + 20° = 110°.

110° — correct
120°
130°
140°