The angle bisector theorem
Answers are for parents and teachers.
1
AD bisects ∠A: AB = 15, AC = 10, BC = 20. Find BD and DC.
BD/DC = 3/2, 5 parts for 20, so BD = 12 and DC = 8.
2
AD bisects ∠A: BD = 6, DC = 10 and the perimeter is 56. Find AB and AC.
BC = 16, AB + AC = 40, AB/AC = 3/5, 8k = 40, k = 5: AB = 15, AC = 25.
3
AB = 14, AC = 10, BC = 12. Find BD, DC (interior bisector) and EC, DE (exterior bisector, E beyond C).
Ratio 7/5: BD = 7, DC = 5. BE = EC + 12 and 5(EC + 12) = 7EC, so EC = 30. DE = DC + CE = 35.
4
AB = 16, AC = 12, BC = 14. Find the length of the bisector AD.
BD = 8, DC = 6, AD² = 192 − 48 = 144, AD = 12.
5
D is on BC. (a) AB = 14, AC = 10, BD = 7, DC = 5: is AD a bisector? (b) AB = 12, AC = 9 and D is the midpoint of BC: is AD a bisector?
(a) 7/5 = 14/10: yes. (b) BD/DC = 1 but AB/AC = 4/3: no.