$D$ is a point inside $\triangle ABC$, $\angle CAD=\angle DAB=10$, $\angle CBD=40$, $\angle DBA=20$, what is $\angle CDB$? $D$ is a point inside $\triangle ABC$, $\angle CAD=\angle DAB=10$, $\angle CBD=40$, $\angle DBA=20$, what is $\angle CDB$?
I'm sure I'm supposed to use trigonometry to obtain the value of $\angle CDB$ but I'm not exactly sure how to start. Maybe Law of Sines?
 A: No need to use trigonometry. Consider a point $X$ on ray $AC$ such that $AB=AX$. Angle chasing quickly leads to $\angle XBD=60^\circ$. Moreover $BD=DX$ as $B$ is symmetric to $X$ with respect to $AD$. So $BDX$ is equilateral and it is easy now to figure out that $\angle BDC=70^\circ$.
A: Let $\measuredangle BDC=x$.
Thus, since $\measuredangle BCD=140^{\circ}-x$ and $\measuredangle DCA=x-40^{\circ},$ by the Cheva's theorem we obtain: 
$$\frac{\sin20^{\circ}\sin(140^{\circ}-x)\sin10^{\circ}}{\sin40^{\circ}\sin(x-40^{\circ})\sin10^{\circ}}=1$$ or
$$\sin(40^{\circ}+x)=2\cos20^{\circ}\sin(x-40^{\circ}),$$ which gives
$$\tan{x}=\frac{\tan40^{\circ}(1+2\cos20^{\circ})}{2\cos20^{\circ}-1},$$
which gives $x=70^{\circ}$ because
$$\frac{\tan40^{\circ}(1+2\cos20^{\circ})}{2\cos20^{\circ}-1}=\frac{\tan40^{\circ}(\cos60^{\circ}+\cos20^{\circ})}{\cos20^{\circ}-\cos60^{\circ}}=$$
$$=\frac{\tan40^{\circ}\cos40^{\circ}\cos20^{\circ}}{\sin20^{\circ}\sin40^{\circ}}=\cot20^{\circ}=\tan70^{\circ}.$$
A: Take E, reflection of B in AD; it belongs to AC and $\triangle BED$ is equilateral; easily BE=BC, BEC being a Langley triangle and required $\widehat{BDC}=70^\circ$.
