In triangle ABC, AB = 9 cm, AC = 6 cm and BC = 5 cm. The bisector of the exterior angle at A meets the line BC produced at E, with C lying between B and E. What is the length of CE?

The external bisector from A divides BC externally in the ratio of the two enclosing sides: EB : $EC$ = AB : AC = 9 : 6 = 3 : 2. Since $C$ lies between B and $E$, EB = $EC$ + BC = $EC+5$, so $(EC+5)/EC=3/2$, giving $2EC+10=3EC$ and $EC=10$ cm. Check: EB = 15 cm and 15 : 10 = 3 : 2. The traps 3 cm and 2 cm are BD and DC for the INTERNAL bisector, which splits the 5 cm side in the same 3 : 2 ratio, and 15 cm is EB rather than the piece $CE$.