Angles of diagonals in a quadrilateral I have a quadrilateral with known angles. Also known is that edge BC and CD have the same length. 
How can I find out the ratio the diagonals divides the angle α into α1 and α2?

 A: I worked this out as I went, so there may be a simpler way, or my equations might simplify to give a direct method nicely, I'm not sure. However, this works:
Since $BC$ and $CD$ have the same length, $\angle CDB=\angle CBD$. We can then use the fact the sum of the three angles of a triangle is $180^\circ$ to determine that exact value. We can then use this knowledge to determine $\angle ADB$ and $\angle ABD$.
Because $|BC|=|CD|$, by the law of cosines, $|DB|^2=2|BC|^2-\cos\gamma$, where $\gamma=\angle BCD$, a known value. Since $\cos(x)\le1<2$, we know that $2|BC|^2-\cos\gamma>0$, so $|DB|=\sqrt{2|BC|^2-\cos\gamma}$.
Now, for convenience, let $\delta=\angle ADB$ and $\beta=\angle ABD$, the values of which are both known from the first part, and let $a=|DB|, b=|AD|, c=|BC|=|CD|$, and $d=|AB|$.
By the law of sines, we know
$$
\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{d}{\sin\delta}
$$
Since, $a, \alpha, \beta,$ and $\delta$ are known, we can determine the values of $b$ and $d$.
Now the law of cosines shows that
$|AC|^2=c^2+d^2-2d\cos(\angle ABC)=c^2+b^2-2b\cos(\angle ADC)$,
so we can determine the length of $AC$.
Applying the law of cosines again, to triangles $ABC$ and $ADC$ gives
\begin{align}
c^2&=d^2+|AC|^2-2d|AC|\cos\alpha_1\\
c^2&=b^2+|AC|^2-2b|AC|\cos\alpha_2\\
\end{align}
So,
\begin{align}
\cos\alpha_1&=\frac{d^2+|AC|^2-c^2}{2d|AC|}\\
\cos\alpha_2&=\frac{b^2+|AC|^2-c^2}{2b|AC|}
\end{align}
Furthermore, since, by the construction of the quadrilateral, $0<\alpha_1<\pi$ and
$0<\alpha_2<\pi$, $\cos^{-1}(\cos\alpha_1)=\alpha_1$ and $\cos^{-1}(\cos\alpha_2)=\alpha_2$
Since you asked about the ratio,
$$
\frac{\cos\alpha_1}{\cos\alpha_2}=
\frac{d^2+|AC|^2-c^2}{b^2+|AC|^2-c^2}
$$
So
$$
\frac{\alpha_1}{\alpha_2}=\cos^{-1}\left(\frac{d^2+|AC|^2-c^2}{b^2+|AC|^2-c^2}\right)
$$
