Ratio of segments in a cube

[samething284]

In a cube ABCDA1B1C1D1, the line AC1 intersects the plane (B, D, A1) at M. Find the ratio, AM: MC1 using vector basis.

So I started this way:
1. Let vector a = AB, b = AD and c = CC1
2. Prove that M is the medicenter of the triangle A1BD
3. Let vector d = (some ratio I can't solve for) = DM because of (2)
4. Express AM and MC1 using a, b, c and d.
5. Find a ratio, then use that a=b=c because it's a cube

 

[samething284]

I solved it.
1 is like in the description
2 - proving M with a new point O in A1BD and prove vector OM=1/3(OD+OB+OA1)
3 - Let K be the continuing of DM, therefore K splits A1B in half, therefore
DK = a - b + 1/2c - 1/2a
Let d = 2/3(DK)
4 - AM = b+d and MC1=-d+a+c, divide both and you should get 1/2
5 - didn't need that