Geometry

let AM = x and BM = y
then MD = 10 - x and CM = 8 - y

Since AM / MD = BM / CM,
x / (10 - x) = y / (8 - y)
8x - xy = 10y - xy
y = 4x / 5

In triangle ABM, by cosine formula,
x^2 = 5^2 + y^2 - 2(5)(y)cos 40
x^2 = 5^2 + (4x / 5)^2 - 2(5)(4x / 5)cos 40
x^2 = 25 + 16x^2 / 25 - (8cos 40)x
25x^2 = 625 + 16x^2 - (200cos 40)x
9x^2 + (200cos 40)x - 625 = 0
x = 3.40 or -20.4(rejected)

y = 4(3.40) / 5 = 2.72

therefore, AM = 3.40 cm and CM = 8 - 2.72 = 5.28 cm

I found the following method workable:
Through A draw a line // to DC and cutting DC produced at E.
Apply cosine rule to triangle ADE to find the length of EC.....