$\Delta ABC$ 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
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 $\gamma=\angle BCA = 30$.