Elementary Geometry The side of the square measures $1\  \mathrm{cm}$ , and $AC = 1\  \mathrm{cm}$, find the value of $AB$

 A: First try:
There are 3 rectangular triangles here, where we get 3 Pythagoras from:
$$
\begin{eqnarray*}
1^2+(1-d)^2&=&c^2\\
a^2+d^2&=&1^2\\
(1+a)^2+1^2&=&(1+c)^2,
\end{eqnarray*}
$$
with $a=\overline{AB}$,$d=\overline{BC}$ and $c=\overline{0C}$, where $0$ is the upper left point. With this, we ask Wolfram to get 
$$
d=\overline{AB}\approx 0.883204 = \frac{1}{2}\left( \sqrt{2}-1+\sqrt{2\sqrt{2}-1} \right)
$$

Second try:
We have $\sin \alpha= \frac{a}{1}= \frac{1+a}{1+c}$ and $1+(1-d)^2=c^2$. 
Further, since $d=\cos( \arcsin (a) )=\sqrt{1-a^2}$, we get
$$
a=\frac{1+a}{1+\sqrt{1+(1-d)^2}}=\frac{1+a}{1+\sqrt{1+(1-\sqrt{1-a^2})^2}}, \tag{*}
$$
with the same solution as above.
$(*)$ can be reformulated to 
$$
a^8-2a^6+7a^4-6a^2+1=A^4-2A^3+7A^2-6A+1=0,
$$
where the roots of polynomial of $4$th order can be found here. But because it works out nice, let's go one step further and substitute $a=u+1/2$, so we get:
$$
u^4+\frac{11}{2}u^2-\frac{7}{16}=0.
$$
From here on, the rest is left to you. Nice Question!
A: Let D and E be the endpoints of the upper side of the square. We know that   $\triangle DEC \cong \triangle ABC. $ Therefore, we have the following: $$EC= \frac{1}{AB+1} ; 1=(1-EC)^2 + AB^2 $$ Solving for $AB$, we have $$ AB^4 + 2AB^3+AB^2-2AB-1 =0$$ Since we know that $0 \le AB \le 1,$ we apply the newton's method by letting $1$ be our initial value in the first iteration. Then we have: $$AB= 0.8832035059$$
A: OK, maybe not a perfect solution, but I managed to transform it into a degree-4 equation.
Let's call the top-left vertex D. Let's call DC x and AB y. Using pythagoras theorem, one can see easily that $(1 + x)^2 = (1 + y)^2 + 1^2$. If we look at the angles CAB and CDE (where E is the top right vertex). They are the same. We can compute their cosine from both angles, once as $1/x$ and the other as $(1+y)/(1+x)$.
If we do the substitution and simplify we get $x^4 + 2x^3 - x^2 -2x - 1 = 0$. Remembering we are trying to compute y.We substitute $y = 1/x$ and get the same answer as draks'.
