Cube question without coordinates, cross sections and planes We have a cube with corners $ABCDA_1B_1C_1D_1$. The points $A_1B_1C_1D_1$ lie above the square $ABCD$. We also have a plane that goes through the middle of line $BC$, through the middle of square $ABA_1B_1$, through the middle of $A_1B_1C_1D_1$. At what ratio does the plane cut line $AB$.
What I did, I've put the cube into the coordinate system $B(0,0,1), A_1 (0,1,0) \text{ and } A(0,0,0)$, $D(1,0,0)$ and then it was pretty easy to solve. I just got the equation of the plane and then solved for intersection of the line that goes through AB, to get the point. Then I knew the ratio.
My question is, how to solve this problem without putting the cube into the coordinate system?
SOLUTION: solution should be that the plane cuts AB in ratio 3:1
Fairly quickly:
(1) the plane (call it $\Pi$) also passes through the centre $M_{ABC_2D_2}$ of square $ABC_2D_2$ ($C_2,D_2$ are the reflection of $C,D$ in line $AB$).  This is just extending $M_{A_1B_1C_1D_1}M_{ABB_1A_1}$ to meet the bottom plane $ABCD$.
(2) So the line $M_{BC}M_{ABC_2D_2}$ lies in $\Pi$.  This line intersects AD at $X$ where $AX=\frac32 AD$ and $X$ is opposite to $D$.
Now look at the configuration $A,B,X,M_{BC}$ and you have similar (right-angled) triangles $\triangle AXY$ and $\triangle YBM_{BC}$ where $Y$ is the intersection of diagonals, i.e., intersection of $AB$ with $\Pi$.  So $AY:YB=AX:BM_{BC}=3:1$.
In response to the question how I solved with coordinate system:
Let's say:
$$AA_1 = (0,0,1) = c$$
$$AB = (0,1,0) = a$$
$$AD = (1,0,0) = b$$
Now we get all three points as mentioned above:
$$P_1 = c + (1/2)a + (1/2)b = (1/2,1/2,1/2)$$
$$P_2 = (1/2)c + (1/2)b = (0,(1/2),(1/2))$$
$$P_3 = a + (1/2)b = ((1/2),1,0)$$
Now we get two vectors $P_2P_3$ and $P_2P_1$. When we do cross product we get: $$n = (1,-2,-1)$$
The equation of the plane is thus: $$x - 2y - z = \frac{-3}{2}$$
Because: $\text{We use the second point and get:}0*1 + (1/2)*(-2)+(1/2)(-1) = \frac{-3}{2}$
Now we know that $AB$ is on  $y$ axis , or the second coordinate as we may call it is a parameter $t$. So the equation of the line on which $AB$ is on is $y = t, x =0, z=0$, where $t$ is the parameter.
Now we input $y =t$ into the plane equation and get $$(-2t = \frac{-3}{2}\iff t =\frac{3}{4} )\implies \text{ratio is: } 1:3$$