# 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!

• I don't know Chinese, but it seems like it says something about $BD$ as well – Andrei Nov 2 '18 at 4:52
• @Andrei It just says "connect $BD$" – Edward Wang Nov 2 '18 at 4:53
• Maybe applying the law of sines will help, but it seems like an ugly equation – Andrei Nov 2 '18 at 5:10

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 $$\pi$$.)
• $180 - 51 - 26 = 103$. That would give a better answer. – Quang Hoang Nov 2 '18 at 5:35
• For the lemma you could also show that $\dfrac{\sin (\alpha-x)}{\sin x}$ is monotonic in that interval – Hari Shankar Nov 2 '18 at 6:03