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 $A$ Given square $ABCD$
I want to construct (with ruler and compass) the circle in the interior of the square such that it is tangent to sides $BC$ and $CD$ and such that it's meetings with the diagonal $BD$ are tangent points from tangents draw from point $A$:

It is clear that the center of the circle must lie in $AC$. I tried finding some cyclic quad somewhere and I failed miserably. I then thought about puting $K$ in hyperbola with focii $A$ and the center $O$ of the square. Then again $K$ lies outside the segment $A O .$
This problem is hard because we would think of looking at the locus of the centers of circles such that the meetings of the circle with line $BD$ are the tangents from $A$.
But that this locus is exactly the same as the locus of the centers of the circles tangent to $CD$ and $BC$: line $AC$.
The proof is simple: as the tangents from $A$ must have the same lenght the meetings $M$ and $N$ of $BD$ with the circles must be reflections of each other with respect to the center $O$ of the square $ABCD$ thus the center of the circle must lie in line $AO$ which is line $AC$.
The real geometric constrain is between the distance of the centers (all of which lie on line $AC$) to point $A$ and the radius of the circles.
Let $P$ be in line segment $OC$.
$PA = x$
$r$ the radius of the circle centered at $P$.
$a=AB$
we have that $r^2 = x^2 - x \frac{a\sqrt2}2$ and $x = a\frac{\sqrt2}4 + \sqrt{r^2+\frac{a^2}8}$
and these weird relations are the "locus" that I desire to work with.
 A: 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.
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. $\square$
