In a triangle ABC AB : AC = 5 : 2, BC = 9 cm. BA is produced to D, and the bisector of the Angle CAD meets BC produced at E. What is the length (in cm) of CE?

9 10 3 6

6

According to the question \(\frac{AB}{AC}\) = \(\frac{BE}{CE}\)  (Using external angle bisector theorem) = \(\frac{5}{2}\) = \(\frac{9 + CE}{CE}\) = 5CE = 18 + 2CE = 3CE = 18 = CE = 6 cm Therefore, CE is 6 cm.