Given a triangle ABC with BC = 2AB. D and E are the midpoints of BC and BD respectively. Show that AD bisects angle CAE.
The Attempt at a Solution
Let AB= x, so DC= AB= x and ED = x/2. If AD bisects angle CAE,
then AC/AE = DC/DE
AC/AE = x/(x/2)
AC: AE = 2:1
That's how far i can do... I tried using sine rule to prove that AC:AE = 2:1 but it did not help. Can anyone guide me?