Find the angle θ (all the circles are tangent) 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]
 A: 
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}
