How to solve for $\angle BDC$ given the information of other angles in the picture 
I want to solve for $\angle BDC$, given $\angle ACB = 26^\circ$, $\angle ABC = 51^\circ$, $\angle BAD = 73^\circ$ and $CD$ bisects $\angle ACB$.
I have tried solving it using the fact that sum of inner angles of a triangle is $180^\circ$, but seems so strange that I cannot get the answer after an hour effort. Thank you for the help!
 A: First, $\angle BAC=180^\circ-\angle ABC-\angle ACB=180^\circ-51^\circ-26^\circ=103^\circ$.  Therefore, $$\angle CAD=\angle BAC-\angle BAD=103^\circ-73^\circ=30^\circ\,.$$  By the trigonometric form of Ceva's Theorem,
$$\left(\frac{\sin(\angle CAD)}{\sin(\angle DAB)}\right)\,\left(\frac{\sin(\angle ABD)}{\sin(\angle DBC)}\right)\,\left(\frac{\sin(\angle BCD)}{\sin(\angle CDA)}\right)=1\,,$$
so
$$\left(\frac{\sin(30^\circ)}{\sin(73^\circ)}\right)\,\left(\frac{\sin(x)}{\sin(51^\circ-x)}\right)\,\left(\frac{\sin(13^\circ)}{\sin(13^\circ)}\right)=1\,,$$
where $x:=\angle ABD$.  This proves that
$$\frac{\sin(x)}{\sin(51^\circ-x)}=\frac{\sin(73^\circ)}{\sin(30^\circ)}=2\,\sin(73^\circ)=2\,\cos(17^\circ)=\frac{\sin(34^\circ)}{\sin(17^\circ)}\,.$$
It follows that $$\frac{\sin(x)}{\sin(51^\circ-x)}=\frac{\sin(34^\circ)}{\sin(51^\circ-34^\circ)}\,.$$
Since $0\leq x\leq 51^\circ$, it is immediate that $x=34^\circ$.  That is,
$$\angle DBC=51^\circ-34^\circ=17^\circ$$
and
$$\angle BDC=180^\circ-17^\circ-13^\circ=150^\circ\,.$$

In the proof above, we use the following lemma.  To be precise, the step where we conclude $x=34^\circ$ follows from this lemma.

Lemma.  If $\alpha$, $\beta$, and $\gamma$ are angles in $(0,\pi)$ such that 
  $$\frac{\sin(\alpha-\beta)}{\sin(\beta)}=\frac{\sin(\alpha-\gamma)}{\sin(\gamma)}\,,$$
  then $\beta=\gamma$.

We have
$$\sin(\beta)\,\sin(\alpha-\gamma)=\sin(\gamma)\,\sin(\alpha-\beta)\,.$$
That is,
$$\cos(\beta+\gamma-\alpha)-\cos(\alpha+\beta-\gamma)=\cos(\beta+\gamma-\alpha)-\cos(\alpha+\gamma-\beta)\,.$$
Consequently,
$$\cos(\alpha+\beta-\gamma)=\cos(\alpha+\gamma-\beta)\,.$$
Hence,
$$\alpha+\beta-\gamma=\pm(\alpha+\gamma-\beta)+2n\pi$$
for some integer $n$.  That is, either
$$\beta-\gamma=n\pi\text{ or }\alpha=n\pi$$
for some integer $n$, but since $0<\alpha<\pi$, we conclude that
$$\beta-\gamma=n\pi$$
for some integer $n$.  Because $-\pi<\beta-\gamma<+\pi$, we must have
$$\beta=\gamma\,.$$  (However, without the restriction that $\alpha,\beta,\gamma\in(0,\pi)$, we can only conclude that either $\alpha$ or $\beta-\gamma$ is an integer multiple of $\pi$.)
