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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!

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

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    $\begingroup$ $180 - 51 - 26 = 103$. That would give a better answer. $\endgroup$ – Quang Hoang Nov 2 '18 at 5:35
  • $\begingroup$ @QuangHoang Ach so! Thanks! $\endgroup$ – Batominovski Nov 2 '18 at 5:37
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    $\begingroup$ For the lemma you could also show that $\dfrac{\sin (\alpha-x)}{\sin x}$ is monotonic in that interval $\endgroup$ – Hari Shankar Nov 2 '18 at 6:03

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