Finding angles plane geometry $\Delta ABC$ is obtuse on $B$ with $\angle ABC = 90 + \frac{\angle BAC}2$ and we have a point $D \in AC$ (in the segment, I mean D is in between A and C) such that $\angle BDA = \angle ABD + \frac{\angle BAC}2$ and $DC = BA$. Find $\angle BCA$.
What I've done:
I've drawn the bissector of $\angle BAC$ and let it meet $BD$ on $F$, then I've got the point $E \in AD$ such that $AE = DC = AB$ therefore we would have $\Delta ABF \equiv \Delta AEF $. The coolest thing I've got from this is that $AF$ is the perpendicular bissector of $BE$ and the triangles $ABC$ and $BEC$ are similar. Believe me or not I've been stuck here for some days ( ._.)
I apreciate the trigonometric solution, if nobody comes with a plane geometry solution I will chose the first answer
 A: Using the information in the problem, you can draw the following diagram:

We can assume that $AB$ has length 1 without loss of generality. Using the information given, we know that
$$ \beta = 90 + \frac{\alpha}{2} $$
$$ \delta = (180-\alpha-\delta) + \frac{\alpha}{2} $$
$$ \alpha+\beta+\gamma=180$$
Also, using the law of sines on the big triangle,
$$ \frac{\sin\beta}{1+x} = \sin \gamma$$
and using the law of sines on the left triangle,
$$ \sin\delta = \frac{\sin (180-\alpha-\delta)}{x}$$
We have 5 equations and 5 unknowns ($\alpha$, $\beta$, $\gamma$, $\delta$,$x$). Eliminating $x$,
$$ \beta = 90 + \frac{\alpha}{2} $$
$$ \delta = (180-\alpha-\delta) + \frac{\alpha}{2} $$
$$ \alpha+\beta+\gamma=180$$
$$ \frac{\sin\beta}{\sin\gamma} = 1+\frac{\sin (\delta-\frac{\alpha}{2})}{\sin\delta}$$
Eliminating $\beta$,
$$ 2\delta = 180- \frac{\alpha}{2} $$
$$ \alpha+90 + \frac{\alpha}{2}+\gamma=180$$
$$ \frac{\sin(90 + \frac{\alpha}{2})}{\sin\gamma} = 1+\frac{\sin (\delta-\frac{\alpha}{2})}{\sin\delta}$$
Eliminating $\delta$,
$$ \alpha+90 + \frac{\alpha}{2}+\gamma=180$$
$$ \frac{\sin(90 + \frac{\alpha}{2})}{\sin\gamma} = 1+\frac{\sin (90- \frac{\alpha}{4}-\frac{\alpha}{2})}{\sin(90- \frac{\alpha}{4})}$$
Simplifying,
$$ 3\alpha+2\gamma=180$$
$$ \frac{\sin(90 + \frac{\alpha}{2})}{\sin\gamma} = 1+\frac{\sin (90- \frac{3}{4}\alpha)}{\sin(90- \frac{\alpha}{4})}$$
Eliminating $\alpha$,
$$ \frac{\sin(90 + \frac{180-2\gamma}{6})}{\sin\gamma} = 1+\frac{\sin (90 - \frac{180-2\gamma}{4})}{\sin(90 - \frac{180-2\gamma}{12})}$$
Solving this gives $\gamma=\angle BCA = 30$.
A: Half way down Victor Liu's solution, I would like to suggest a different way to proceed
$\frac{\sin  \beta }{\sin  \gamma }=1+\frac{\sin (\delta -\alpha /2)}{\sin  \delta }=\frac{\sin  \delta +\sin (\delta -\alpha /2)}{\sin  \delta }$
$\frac{\sin  \beta }{\sin  \gamma }=\frac{2\sin (\delta -\alpha /4)\cos (\alpha /4)}{\sin  \delta }$
$\sin  \delta =\cos (\alpha /4)$
$\sin  \beta =\sin (\delta -\alpha /4)$
$2\sin  \gamma =1$
