Vectors

a) AB = AD + DB = -2a + 6 b
DM = (DB + DA) / 2 = (6b + 2a)/2 = 3b + a

b) i) BC = kDM
BC = k(3b + a) = ka + 3kb
AC = 2DB
AC = 2(6b) = 12b
AB = -2a + 6b
AB + BC = AC
-2a + 6b + ka + 3kb = 12b
ka + 3kb = 2a + 6b
k(a + 3b) = 2(a + 3b)
k = 2

ii) BC = 2DM
Thus BC // DM, i.e. BC // DN.
And AM = MB
By intercept theorem,
AN = (1/2)AC