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Question given below:

question

ABC is a triangle and D is a point inside ABC such that:

$$ m(\widehat{DCB})=m(\widehat{CBD})=18^{\circ}\\ m(\widehat{ACD})=24^{\circ}\\ m(\widehat{DBA})=12^{\circ}\\ m(\widehat{DAC})=\alpha=? $$

This is supposed to be a high-school level question. But i can't find the α. Is there something obvious i'm missing?

I checked that question is well-posed and α = 78 degrees. If i can just show that $|CD|=|CA|$ then i'm done. But that's the only step i can take in this question (If this can be called a step at all.).

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  • $\begingroup$ Are you sure there is no more info about the point $D$? Like it being circumcenter or something? $\endgroup$ – taninamdar Oct 2 '14 at 15:26
  • $\begingroup$ what else is given? $\endgroup$ – Dr. Sonnhard Graubner Oct 2 '14 at 15:31
  • $\begingroup$ @taninamdar Nope. That's all given. No more information other than that. Also, i checked now (out of curiosity) and D is not circumcenter in this question. $\endgroup$ – Alistair Oct 2 '14 at 15:35
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By the Trig Ceva Theorem, $$\frac{\sin(12^\circ)}{\sin(18^\circ)}\cdot \frac{\sin(18^\circ)}{\sin(24^\circ)}\cdot\frac{\sin(\alpha)}{\sin(108^\circ-\alpha)}=1$$ hence: $$\sin(96^\circ-\alpha)+\sin(120^\circ-\alpha)=2\cos(12^\circ)\sin(108^\circ-\alpha)=\sin(\alpha)$$ or: $$2\cos(12^\circ)\sin(108^\circ)\cos(\alpha)=(2\cos(12^\circ)\cos(108^\circ)+1)\sin(\alpha)$$ or: $$\tan(\alpha)=\frac{2\cos(12^\circ)\sin(108^\circ)}{2\cos(12^\circ)\cos(108^\circ)+1}$$ from which it is tedious (but not difficult) to check that $\alpha=78^\circ$ is the wanted solution.

The identity: $$\sin(18^\circ)+\sin(42^\circ)=\sin(78^\circ)=\cos(12^\circ)$$ is interesting in itself, but just follows from the Briggs formulas.

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  • $\begingroup$ Thanks for the answer. But i was looking for more elementary methods. I was hoping maybe i can reduce your answer to that, but nothing comes up. Do you have any idea how can i find an elementary solution to this question? $\endgroup$ – Alistair Oct 2 '14 at 16:05
  • $\begingroup$ @Alistair: Honestly I don't know. It is sufficient to prove that $DAC$ is an isosceles triangle, or that the angle bisector of $\widehat{DAC}$ is perpendicular to $AD$, but it isn't so obvious to me at the first glance. $\endgroup$ – Jack D'Aurizio Oct 2 '14 at 17:43
  • $\begingroup$ Okay. But, in your $\tan\alpha$ formula, how am i going to find out that $\alpha=78^\circ$ without calculator? It seems all things come down to a comparison between algebraic representation of $\tan78^\circ$ and some formula involving (at least) algebraic representation of $\sin6^\circ$ (or something close to it). $\endgroup$ – Alistair Oct 2 '14 at 20:35
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I solved this with elementary methods. In case of anyone interested comes and looking for answer and for archiving reasons, i post the solution:

Solution

Draw angle bisector of $\widehat{ACD}$ making $\overset{\Delta}{BEC}$ isosceles. Draw perpendicular from $E$ to $[BC]$ trisecting $\widehat{BEA}$ and making $\overset{\square}{ACDE}$ kite, therefore proving $|CA|=|CD|$.

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