Week 10 · Angle bisectors and proportional parts: interior and exterior bisectors and the length of a bisector · 3 lessons
3
The length of a bisector, AD² = AB × AC − BD × DC, the converse of the theorem and the bisectors that meet at one point
Mathematics Sec 1 Term 1, Unit 3, Lesson 2
GoalFind the length of an angle bisector and use the converse of the bisector theorem.
New words
length of the bisector · AD² = AB × AC − BD × DCconcurrent lines · lines that all pass through one point
Explanation
Length of the bisector: if AD bisects ∠A with D on BC, then AD² = AB × AC − BD × DC. (The proof uses the circle through A, B, C and the chord theorem of Week 8: AD × DE = BD × DC, where E is where the bisector meets the circle again.) Test it on a triangle with AB = 21, AC = 14, BC = 25: BD/DC = 3/2, so BD = 15 and DC = 10, and AD² = 294 − 150 = 144, AD = 12. The converse: if D is on BC and BD/DC = AB/AC, then AD bisects ∠A; check the ratio with the whole numbers. Same for an outside point E and the exterior bisector. The three bisectors of the angles of a triangle meet at one point, the centre of the circle that touches all three sides. The formula for the length of the exterior bisector is AE² = BE × EC − AB × AC.
ƒFormula
AD² = AB × AC − BD × DC
the square of the bisector = the product of the two sides − the product of the two parts
Worked examples
AB = 21, AC = 14, BC = 25 and AD bisects ∠A. BD = 15, DC = 10. AD² = 21 × 14 − 15 × 10 = 294 − 150 = 144, so AD = 12 cm.
AB = 14, AC = 10 and BC = 12, with D on BC, BD = 7 and DC = 5. Is AD the bisector of ∠A? BD/DC = 7/5 = 14/10 = AB/AC, so yes by the converse. Then AD² = 140 − 35 = 105.
AD is a median of △ABC (D is the midpoint of BC). DX bisects ∠ADB with X on AB and DY bisects ∠ADC with Y on AC. AX/XB = AD/DB and AY/YC = AD/DC. Since DB = DC, AX/XB = AY/YC, so XY ∥ BC.
Class activity
Measure and check. Draw a triangle with AB = 8 cm, AC = 6 cm, BC = 7 cm. Construct the bisector of the angle at A with a compass and measure AD. Compute AD² = AB × AC − BD × DC and compare.
Practice
1
AB = 16, AC = 12, BC = 14, AD bisects ∠A. Find AD.
BD/DC = 4/3: BD = 8, DC = 6. AD² = 192 − 48 = 144, AD = 12.
2
For AB = 12, AC = 8, BC = 10 find the length of the interior bisector AD.
BD = 6, DC = 4. AD² = 96 − 24 = 72, AD = 6√2.
3
In the same triangle find the length AE of the exterior bisector.
