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The problem:

enter image description here

Answer:

Extend $AC$ to the side of $C$, such that $BC=CE$. Then, construct the equilateral triangle $BEF$. Triangles $CAB$ and $EFC$ are congruent, thus $AC=BE$. Now, observe that $AC=DE$, thus $BE=DE$. It follows that $\triangle BED$ is isosceles, thus $\angle EDB=80$.

enter image description here

I was given this problem. The provided construction is smart, but I have a feeling that there should be a simpler non-trigonometric approach. Can anyone think of one?

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  • $\begingroup$ Well, I have a solution which also use a constructon of an equilateral triangle. I will post a solution tommorow if you are interested. $\endgroup$ – Aqua Apr 6 at 22:12
  • $\begingroup$ @Maria Sure. Please do. $\endgroup$ – blackened Apr 7 at 2:14
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Say $O$ is a circumcenter of circumcircle $(ABC)$ and let $BO$ cut $AC$ at $E$. Then $$\angle COB = 2 \angle CAB = 60$$ so triangle $BCO$ is equaliteral. So we have $$ BC =OC=OB = OA$$ Since $$\angle AOB = 2 \angle ACB = 80$$ and since $\angle OAE = 20$ (note that $ACO$ is isosceles) we have $\angle AEO = 80$ so $AO=AE$ which means $E =D$, and finaly we have $\angle CDB =80$. enter image description here

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  • $\begingroup$ Thanks. This is also a clever solution. But would you assert that it is simple? One first has to forget about point $D$ (and thus the fact that $AC=BE$), do some auxiliary constructions, then show that ${E}={D}$. $\endgroup$ – blackened Apr 7 at 8:18
  • $\begingroup$ Probably not, but then again I suppose there is no simple way for this problem. However, construction of the circumcircle and circumcenter should not be so difficult. What do you think? $\endgroup$ – Aqua Apr 7 at 9:55

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