Angle Between Two Planes (No Vector and Coordinates Allowed, Only Allowed to Use Non-Coordinate Geometry Tricks) Cube $ABCD.EFGH$ have side length 6 units. P is the midpoint of $EH$, Q is the midpoint of $AD$, the angle between $BFPQ$ and $BDG$ is $\lambda$.
My attempt :
1) Because the line of intersection is outside the cube, i translated $BFPQ$ 3 units forward (from my point of view) so it become $DHP'Q'$ and the line intersection after translation is $DO$.
2) I calculated the distance from $B$ to $DO$ and distance from $Q'$ to $DO$, because in order to find the angle between two plane (without relying on normal vector) you must find 2 lines perpendicular to the line of intersection, Those 2 perpendicular lines in this case are $BO$ and $Q'U$
3) When i checked in GeoGebra it turns out that $BO$ and $Q'U$ are skew to each other. That's what caused my answer to be wrong.
My question is : How do i avoid the 2 lines perpendicular to the lines of intersection being skew to each other (avoid $BO$ and $Q'U$ being skew) ? and if i have to translate the skew lines how do i figure out how far do i need to translate one of the line so that the two lines aren't skew anymore ? is there any other easier way to do this ? of course without vector and coordinate.

 A: The two lines must be perpendicular to the line of intersection at the same point. In your case, the simplest thing to do is drawing a perpendicular $UV$ on plane $BDG$ (I don't know if point $V$ in your figure is the one I mean) and compute $\lambda=\angle Q'UV$.
EDIT.
Computing $UV$ and $Q'V$ is not difficult:


*

*in rectangle $DQ'P'H$ you have $UQ'=3\sqrt{\frac{5}{6}}$ and $DU=\frac{5\sqrt6}{2}=\frac{5}{6}DO$; 

*in equilateral triangle $DBG$ line $UV$ is parallel to $BG$, hence 
$DV=\frac{5}{6}DB=5\sqrt2$ and $UV=\frac{5}{2}\sqrt2$.
A: It’s actually pretty straightforward. 
Let $X$ be the midpoint of $AB$. Then, $ΔCEX$ is an isosceles triangle with side lengths 
$$\frac{\sqrt{5}}{2}, \space \frac{\sqrt{5}}{2}, \space \sqrt{3},$$ 
assuming a unit cube.
The angle between the two planes is just $\lambda=\angle{ECX}$, because $CE$ is normal to the plane $BDG$ due to $xyz$-symmetry and $CX$ is normal to $BFPQ$ due to $CX⊥BQ$ and $BFPQ$‘s vertical orientation.
Thus, from the dimensions of the isosceles triangle given above,
$$\cos\lambda = \sqrt{\frac{3}{5}}$$
