In the following figure ABCD is a side square $\alpha$, the points $P_0, P_1, P_2, P_3, Q_0, Q_1, Q_2, Q_3, X \ and \ Y$ are points of tangency, $BC \ and \ ZB$ are the diameters, respectively, of the blue and green semi-circles. Determine the angle $\theta$ Answer:$θ=67,5°$
There is a lot of homoteties, but I only could find that LK=$\frac{\sqrtα}{4}$. I guess that $BP_1$ are diagonal of the square, but I don't know how to prove (or disprove) this.
Can someone help me to solve this problem?
Thanks for antetion. 
[Question image]
Let $\angle P_1BC=\phi=180^\circ-2\theta$.
Next, let's find $r_0$ and $r_1$.
\begin{align} \triangle LO_0E:& \\ |O_0L|&=r+r_0=\tfrac a2+r_0 ,\quad |LE|=\tfrac a2-r_0 ,\\ |O_0E|=|BX|&= \sqrt{2 a r_0} \tag{1}\label{1} ,\\ \triangle BO_0X:& \\ |BO_0|&=R-r_0=a-r_0 ,\quad |O_0X|=r_0 ,\\ |BX|&=\sqrt{a^2-2 a r_0} \tag{2}\label{2} . \end{align}
Since \eqref{1}=\eqref{2}, we have
\begin{align} r_0&=\frac a4 . \end{align}
By Descartes' theorem for four mutually tangent circles with radii $R,r,r_0$ and $r_1$,
\begin{align} r_1&= \left( \tfrac1r+\tfrac1{r_0}-\tfrac1R +2\sqrt{\tfrac1{r r_0}-\tfrac1{r R}-\tfrac1{r_0 R}} \right)^{-1} \\ &= \frac{5-2\sqrt2}{17}\,a . \end{align}
Now, consider a Steiner’s chain of circles with two reference circles: external, centered at $B$ with radius $R=a$ and internal, with the center at $L$ and the radius $r=\tfrac a2$, and known distance between the centers, $d=|BL|=r=\tfrac a2$.
Ignoring all the rest, let's assume that the circle $O_1$ with already known radius $r_1$, is the first in this Steiner’s chain of circles. Then we can exploit a known formula to find the angle $\phi$:
\begin{align} r_1&=R-\tfrac12\,\frac{(r+R)^2-d^2}{r+R-d\,\cos\phi} ,\\ \cos\phi&= \frac{a-3r_1}{a-r_1} = \frac{a-3\frac{5-2\sqrt2}{17}\,a}{a-\frac{5-2\sqrt2}{17}\,a} =\frac{\sqrt2}2 , \end{align}
and the answer follows.
Edit
Alternatively, the same result can be obtained much simpler: since all of the sides of $\triangle BLO_1$ are known, we can just use the cosine law:
\begin{align} \cos\phi&= \frac{|BO_1|^2+|BL|^2-|LO_1|^2}{2\cdot|BO_1|\cdot|BL|} = \frac{(R-r_1)^2+\tfrac{a^2}4-(r-r_1)^2}{2(R-r_1)^2\cdot\tfrac{a^2}4} =\tfrac{\sqrt2}2 . \end{align}