Angles of diagonals in a quadrilateral

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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?

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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 $$ \frac{\alpha_1}{\alpha_2}=\cos^{-1}\left(\frac{d^2+|AC|^2-c^2}{b^2+|AC|^2-c^2}\right) $$