Calculate the angle $x$, no other information was given

Question

I tried to solve this question, looked at this for very long and still was not able to find a solution. Can anybody here help me. This question was asked by our teacher and I have no information from where he got it. []

Answer 1

$$\begin{align} \tan x = &\frac{AE}{BE} =\frac{AE}{DE}\cdot \frac{CE}{BE}\cdot \frac{DE}{CE} =\tan70\tan80\tan60\\ =&\frac{\sin70\cos80\sin60}{\cos40\sin40\cos70 } =\frac{\sin70(\sin40-\sin20)}{\cos40(\cos20+\cos120) } =\frac{\sin70\sin40-\frac12 \sin40}{\cos40(\sin70-\frac12) }\\ =& \frac{\sin40}{\cos40 }=\tan40\\ \end{align}$$

Answer 2

Let $E$ be the circumcenter of triangle $BCD$. Then $\angle BEC = 2\angle BDC = 60^\circ$ and additionally $EB=EC$, hence $EBC$ is equilateral. Also note that $\angle CAD = 20^\circ =2\angle CBD =\angle CED$, hence $EADC$ is cyclic. Moreover, $\angle EDB =90^\circ -\frac 12 \angle BED = 90^\circ -\frac 12 80^\circ = 50^\circ$, so $\angle ADE =\angle ADB -\angle EDB =70^\circ -50^\circ =20^\circ =\angle CED$. It follows that $EC \parallel AD$. So $EADC$ is a cyclic trapezoid, hence it is isosceles and now we see that our picture is symmetric with respect to perpendicular bisector of $CE$. It is easy to calculate now that $\angle DBA = 40^\circ$.