Construct a circle tangent to sides $BC$ and $CD$ and s.t. its meetings with the diagonal $BD$ are tangent points from tangents draw from point $A$

Let center $O$ of the circle lie on diagonal $\overline{AB}$ with midpoint $M$, and define $a:=|OA|$, $b:=|OB|$. Let the circle meet the other diagonal at $R$, and define $r:=|OR|$; note that $r=b/\sqrt{2}$.

$$\begin{align} \underbrace{\frac{|OR|}{|OA|}=\frac{|OM|}{|OR|}}_{\triangle ORA\sim\triangle OMR} &\quad\to\quad \frac{r}{a}=\frac{a-\frac12(a+b)}{r} =\frac{a-b}{2r}\tag{1} \\ &\quad\to\quad a(a-b)=2r^2=b^2 \tag{2} \\[8pt] &\quad\to\quad \frac{a}{b}=\frac{b}{a-b}=\phi \tag{3} \end{align}$$ (ignoring a negative solution) where $\phi := \frac12(1+\sqrt{5})$ is the Golden Ratio.

Consequently, the construction reduces to dividing diagonal $\overline{AB}$ in the ratio $\phi:1$. A simple method for doing so is described under "Dividing a line segment by interior division" in the Wikipedia entry. $\square$

Just a hint: If $K$ is the center of the circle, $O$ is the center of the square and $L$ is the tangency point lying on $BD$ then triangles $KOL$ and $KLA$ are similar. This gives $KO\cdot KA=KL^2$. This allows you to calculate the radius of the circle.