Basic vector algebra suffices here!
[Rewritten on 2016-06-01, using more general solution with the same "cost", number of multiplications, additions, etc.]
Let's assume the blue parallelogram is your bounding box, and the red one (on the right) is the test rectangle -- or, really, any set of points you might wish to test:
You pick one corner of the parallelogram as its origin ($\vec{b}_0 = (x_0, y_0)$). We don't use the point diagonally opposite. We call one edge $\vec{n}_u = (x_u, y_u)$, and another $\vec{n}_v = (x_v, y_v)$; both start at the $\vec{b}_0 = (x_0, y_0)$ point.
Note that this means the other points on the parallelogram are
$$\vec{b}_w = \vec{b}_0 + \vec{n}_u = (x_0 + x_u, \; y_0 + y_u) \\
\vec{b}_h = \vec{b}_0 + \vec{n}_v = (x_0 + x_v, \; y_0 + y_v)$$
and the unmarked fourth point is at $(x_0 + x_u + x_v, \; y_0 + y_u + y_v)$.
We can transform to a coordinate system that is rotated, translated, scaled, and skewed so that the original parallelogram is at coordinates $(0,0)$-$(1,0)$-$(1,1)$-$(0,1)$. Let's call these coordinates $(u,v)$.
To convert back to ordinary coordinates we use
$$\begin{cases}
x = x_0 + u x_u + v x_v \\
y = y_0 + u y_u + v y_v
\end{cases}$$
Solving the above for $u$ and $v$ gives us the interesting conversion to the $(u,v)$ coordinates:
$$\begin{cases}
u = \frac{(x - x_0)y_v - (y - y_0)x_v}{x_u y_v - y_u x_v} \\
v = \frac{(y - y_0)x_u - (x - x_0)y_u}{x_u y_v - y_u x_v} \\
\end{cases}$$
Combining the conversion with the limits shown earlier, we can condense the test. Do note that the divisor (or limit),
$$L = x_u y_v - y_u x_v$$
may be positive or negative. A point $(x,y)$ is inside our bounding parallelogram if and only if
$$L \gt 0: \begin{cases}
0 \le (x - x_0)y_v - (y - y_0)x_v \le L \\
0 \le (y - y_0)x_u - (x - x_0)y_u \le L
\end{cases} \\
L \lt 0: \begin{cases}
0 \le (y - y_0)x_v - (x - x_0)y_v \le -L \\
0 \le (x - x_0)y_u - (y - y_0)x_u \le -L
\end{cases}$$
$L \ne 0$ unless one or both edges are zero length, or if they are parallel. In fact, $L$ is actually the area of the parallelogram; zero $L$ means the parallelogram has zero area (vanishes).
Now, I originally had a "simpler" test using direct projection (dot product, also known as projection product), but it only works if the edges are perpendicular to each other. This one works for squares, rectangles, and parallelograms -- and has exactly the same number of multiplications and additions; only the sign check is extra -- so this one is clearly better choice.
Let's look at OP's example bounding box, with corners at $(1,6)$, $(-6,-5)$, $(2,-9)$, and $(9,3)$. I'll choose, arbitrarily,
$$\begin{cases}
\vec{b}_0 = (-6, -5) \\
\vec{b}_w = (1, 6) \\
\vec{b}_h = (2, -9)
\end{cases} \iff \left\lbrace\begin{array}{ll}
x_0 = -6, & y_0 = -5 \\
x_u = 7, & y_u = 11 \\
x_v = 8, & y_v = -4 \end{array}\right.$$
This means that the fourth point is actually $(-6+7+8,-5+11-4) = (9,2)$, and not $(9,3)$; OP's bounding region is a quadrilateral, not a parallelogram. (Because $\vec{n}_u\cdot\vec{n}_v = 7\cdot8-11\cdot4 = 12 \ne 0$, the edges are not perpendicular; the box is not rectangular either way.)
So, assuming the fourth corner of the bounding parallelogram is at $(9,2)$, we can test for some points. We have
$$L = 7(-4) - 8(11) = -116$$and therefore the inclusion tests are
$$\begin{cases}
0 \; \le \; (y + 5)8 + (x + 6)4 \; \le \; 116\\
0 \; \le \; (x + 6)11 - (y + 5)7 \; \le \; 116
\end{cases}$$
Let's check point $(x,y) = (12, 16)$ first. (We know it has to be outside.)
$$\begin{cases}
(y+5)8 + (x+6)4 = (16+5)8 + (12+6)4 = 240 \\
(x+6)11 - (y+5)7 = (12+6)11 - (16+5)7 = 51
\end{cases}$$
the first is greater than $116$, so the point is definitely outside.
Let's check point $(x,y) = (3,3)$. Remember, OP's diagram has a quadrilateral, which should be fixed to put the fourth corner at $(9,2)$, to make it a quadrilateral.
$$\begin{cases}
(y+5)8 + (x+6)4 = (3+5)8 + (3+6)4 = 100 \\
(x+6)11 - (y+5)7 = (3+6)11 - (3+5)7 = 43
\end{cases}$$
But yes, point $(3,3)$ is inside the fixed quadrilateral, too.
If you have an arbitrary quadrilateral, an arbitrary polygon (convex or concave), there are other methods one can do to determine whether a point is inside it or not. Many of these cases have been asked and answered before, and each of them have different efficient numerical solutions, so combining them all into a single answer is not sensible.
Here is some pseudocode to help with implementation.
Let's assume we have variables
x0,y0 # b0, the origin corner of the bounding parallelogram
xw,yw # bw, another corner of the bounding parallelogram
xh,yh # bh, the third corner of the bounding parallelogram
Each time the above change, I would precalculate five variables:
xu = xw - x0
yu = yw - y0
xv = xh - x0
yv = yh - y0
L = xu * yv - xv * yu
if L < 0:
L = -L
xu = -xu
yv = -yv
else:
xv = -xv
yu = -yu
end if
The above negates the constants when necessary, so that only one case, one set of tests, is left: 0 <= (x-x0)*yv + (y-y0)*xv <= L and 0 <= (x-x0)*yu + (y-y0)*xu <= L. This means the test for each point (x,y) is
u = (x - x0)*yv + (y - y0)*xv
if (u < 0 || u > L):
return OUTSIDE
else:
v = (x - x0)*yu + (y - y0)*xu
if (v < 0 || v > L):
return OUTSIDE
else:
return INSIDE
end if
end if
which boils down to four multiplications, four additions and two subtractions, and four comparisons, maximum, to determine point inclusion in the parallelogram.
Best Answer
Let $X=(0,0)$ and let $W=(a,b)$. Call $S_X, S_W$ the respective squares, $S_X', S_W'$ after rotation.
$$S_X : (a\pm EF/2, b\pm EF/2)$$
$$S_W : (\pm AB/2, \pm AB/2)$$
After rotation:
$$S_X' : (a, b\pm \sqrt{2}/2EF),(a\pm \sqrt{2}/2EF, b)$$
$$S_W' : (0, \pm \sqrt{2}/2AB), (\pm \sqrt{2}/2AB,0)$$
(you can get this by multiplying the coordinates by the rotation matrix.)
If you want to line up $B$ and $F$ then you want real numbers $s,t$ such that
$$(s, \sqrt{2}/2AB+t)=(a, b+ \sqrt{2}/2EF)$$
So $$(s,t)= (a,b+\sqrt{2}/2(EF-AB))$$
Edit
Here is an interactive plot I made for you. First both squares of specified length are rotated by 45 degrees. Then the square centered at the origin is translated by the amount specified above.
For rectangles, rather than go through a whole 'nother derivation, the value of $s,t$ you seek is:
$$\left[\begin{matrix}s\\ t\end{matrix}\right]=\left[\begin{matrix}a\\ b\end{matrix}\right]+\left[\begin{matrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{matrix}\right]\left[\begin{matrix}F_x-B_x-a\\ F_y-B_y-b\end{matrix}\right]$$
Another example for rectangles, rotating $270^o$