7 min readConcept sheetPart III proof or problem (5 marks); Part II (2 marks)

Angle bisector theorem and its converse

An angle bisector of a triangle splits the opposite side in the ratio of the two sides beside the angle: BD/DC = AB/AC.

Angle bisector theoremIf AD bisects ∠A of ΔABC, then BD/DC = AB/AC

ConverseIf BD/DC = AB/AC, then AD bisects ∠A

Finding the partsBD = BC × AB/(AB + AC)

Proof ideaDraw CE ∥ AB: ΔACE is isosceles and ΔABD ~ ΔECD

Theorem vs converse

Angle bisector theoremThe internal bisector of an angle divides the opposite side internally in the ratio of the sides containing the angle. So BD/DC = AB/AC.

ConverseA line through a vertex divides the opposite side in the ratio of the other two sides. Then it bisects the angle at that vertex.

A bisector is a line that cuts an angle into two equal parts. The side next to B goes over the side next to C: BD over DC, AB over AC.

How to draw the figure and prove the theorem

Figure: draw ΔABC. Draw AD bisecting ∠A, meeting BC at D. Through C, draw CE parallel to AB. Extend AD to meet CE at E. Mark the equal angles.

1
Given: in ΔABC, AD bisects ∠BAC and meets BC at D. To prove: AB/AC = BD/DC.
2
Construction: through C, draw CE ∥ AB. Extend AD to meet CE at E.
3
AB ∥ CE and AE cuts both, so ∠CEA = ∠BAE.
alternate angles
4
But ∠BAE = ∠CAE, so ∠CEA = ∠CAE.
AD bisects ∠A
5
In ΔACE, two angles are equal, so the sides opposite them are equal: CE = AC.
ΔACE is isosceles
6
In ΔABD and ΔECD: ∠BAD = ∠CED and ∠ADB = ∠EDC. So ΔABD ~ ΔECD.
vertically opposite angles; AA
7
So AB/CE = BD/DC. Put CE = AC: AB/AC = BD/DC. Hence proved.

Example 1: find the two parts

Problem

In ΔABC, AB = 6 cm, AC = 8 cm and BC = 7 cm. The bisector of ∠A meets BC at D. Find BD and DC.

Working

By the angle bisector theorem, BD/DC = AB/AC = 6/8 = 3/4.
So BC is split in the ratio 3 : 4. Total parts = 7.
BD = (3/7) × 7 = 3 cm
DC = (4/7) × 7 = 4 cm

Answer

BD = 3 cm and DC = 4 cm

Example 2: a path across a field (word problem)

Problem

A triangular field ABC has AB = 120 m, AC = 80 m and BC = 100 m. A straight path from A bisects ∠A and meets BC at D. How far is D from B?

Working

The path bisects ∠A, so BD/DC = AB/AC = 120/80 = 3/2.
BC = 100 m is split in the ratio 3 : 2. Total parts = 5.
BD = (3/5) × 100 = 60 m
DC = 100 − 60 = 40 m

Answer

D is 60 m from B (and 40 m from C).

Example 3: use the converse

Problem

In ΔPQR, PQ = 5 cm and PR = 7.5 cm. S is on QR with QS = 2 cm and SR = 3 cm. Does PS bisect ∠P?

Working

PQ/PR = 5/7.5 = 2/3
QS/SR = 2/3
QS/SR = PQ/PR, so by the converse of the angle bisector theorem, PS bisects ∠P.

Answer

Yes, PS bisects ∠QPR.

In real life

When a straight path bisects the corner of a triangular plot, this theorem tells exactly where it meets the far boundary. Surveyors use such ratios to divide land fairly.

Exam angle

2 marks: “In ΔABC, AD bisects ∠A. AB = 8 cm, AC = 12 cm and BC = 7.5 cm. Find BD.” BD/DC = 8/12 = 2/3. Total parts = 5, so BD = (2/5) × 7.5 = 3 cm.

5 marks: “State and prove the angle bisector theorem.” Statement and figure with CE ∥ AB and AD extended to E. Alternate angles give CE = AC. ΔABD ~ ΔECD by AA, so AB/CE = BD/DC. Replace CE by AC.

2 marks: “In ΔPQR, PQ = 6 cm and PR = 9 cm. S is on QR with QS = 4 cm and SR = 6 cm. Is PS the bisector of ∠P?” PQ/PR = 6/9 = 2/3 and QS/SR = 4/6 = 2/3. The ratios are equal, so PS bisects ∠P by the converse.

Remember it

The bisector splits the base like the arms: BD : DC = AB : AC.