Difficult geometry question involving Pythagoras theorem? 
Hello mathematicians,
I was given this question by my teacher and after spending a couple of hours looking over it have not been able to solve it. 
I understand it involves radians which I have attempted to learn through a few quick online courses, which have only helped me in understanding:
$\frac{3*\pi}4 = 135°$
Which is not helpful in solving the question.
The annoying thing is this is in a chapter titled Pythagoras Theorem and it Converse. Yet I am having a lot of trouble figuring it out! As always the textbook does not include answers as it is a show question.
Any pointers are appreciated.
 A: Here is something to get you started:
Place a point $Q$ on the segment $AB$ such that $AB$ and $PQ$ meet at right angles:

Let us introduce the variables $s=|AD|,x=|AQ|,y=|PQ|$. Then $s$ is the side length of the square. We then must have
$$
\begin{align}
x^2+y^2&=4\\
(s-y)^2+x^2&=1\\
(s-x)^2+y^2&=9
\end{align}
$$
From the first equation we get $x^2=4-y^2$, and substituting this into the second equation we can solve for $y$ to have $y=\frac{s^2+3}{2s}$. Similarly, we may substitute $y^2=4-x^2$ into the third equation and solve for $x$ to have $x=\frac{s^2-5}{2s}$. With this, the first equation becomes
$$
\left(\frac{s^2-5}{2s}\right)^2+\left(\frac{s^2+3}{2s}\right)^2=4
$$
which after multiplying both sides by $4s^2$ and using the substitution $t=s^2$ yields the quadratic equation
$$
(t-5)^2+(t+3)^2=16t\iff s^2=t=5\pm2\sqrt 2
$$
Now, since $x=\frac{s^2-5}{2s}>0$ it must be $s^2=5+2\sqrt 2$. Therefore we get
$$
\begin{align}
\cos\theta&=\frac{1^2+2^2-s^2}{2\cdot 1\cdot 2}\\
&=-\frac{\sqrt 2}2
\end{align}
$$
which can easily be derived or found in a table to be equivalent to $\theta=\frac{3\pi}4$.
A: So turns out it is possible to do this question with Pythagoras theorem alone!
I found the proof online here

However I do not imagine my self ever coming up with this in an exam. 
Regardless thank you String and Archaick for helping me out.
