In a triangle ABC, AB=2 cm, BC=4 cm and AC=3 cm. The bisector of angle A meets BC at D and the bisector of angle B meets AD at E. What is AE:ED equal to?

Explanation

By Angle Bisector Theorem in \triangle ABC, BD = \frac{AB}{AB+AC} \times BC = \frac{2}{5} \times 4 = 1.6. In \triangle ABD, the angle bisector BE gives \frac{AE}{ED} = \frac{AB}{BD} = \frac{2}{1.6} = 5:4.

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