In triangle ABC, AB = 12 cm, ∠B = 60°, the perpendicular from A to BC meets it at D. The bisector of ∠ABC meets AD at E. Then E divides AD in the ratio

Given : $$\angle$$ABC = 60 , AB = 12 cm

To find : AE : ED

Solution : From $$\triangle$$ABD

=> $$sin 60 = \frac{AD}{BD}$$

=> $$\frac{\sqrt{3}}{2} = \frac{AD}{12}$$

=> $$AD = 6\sqrt{3}$$ cm

Again,

=> $$cos 60 = \frac{BD}{AB}$$

=> $$\frac{1}{2} = \frac{BD}{12}$$

=> $$BD = 6$$ cm

Also, BF is angle bisector of angle B => $$\angle$$EBD = 30

From $$\triangle$$ BDE

=> $$tan 30 = \frac{DE}{BD}$$

=> $$\frac{1}{\sqrt{3}} = \frac{DE}{6}$$

=> $$DE = 2\sqrt{3}$$ cm

$$\therefore$$ $$\frac{AE}{ED} = \frac{AD - ED}{ED}$$

= $$\frac{6\sqrt{3} - 2\sqrt{3}}{2\sqrt{3}}$$

= $$\frac{4\sqrt{3}}{2\sqrt{3}} = \frac{2}{1}$$

=> Required ratio = 2 : 1