I have to find the value of $x$ and I did it and found out $x=40^0$. I used the cot theorem to find it. but without trigonometry is there a purely geometric proof? every single time I tried I was stopped at $x+F\hat BD=80^0$. Any hint or a proof would be appreciated.
P.S this is different from Langley’s Adventitious Angles problem since $CE \neq AC$ I saw some basic constructions led me to have some cyclic quadrilaterals. but they also did not gave me the answer
On
Easy to see that $$\measuredangle DBC=20^{\circ}+x$$ and $$\measuredangle EBD=80^{\circ}-x.$$ Thus, by law of sines we obtain: $$\frac{ED}{DC}=\frac{\frac{ED}{BD}}{\frac{DC}{BD}}=\frac{\frac{\sin(80^{\circ}-x)}{\sin30^{\circ}}}{\frac{\sin(20^{\circ}+x)}{\sin50^{\circ}}}=\frac{2\sin(80^{\circ}-x)\sin50^{\circ}}{\sin(20^{\circ}+x)}.$$ In another hand, $$\frac{ED}{DC}=\frac{\frac{ED}{AD}}{\frac{DC}{AD}}=\frac{\frac{\sin30^{\circ}}{\sin80^{\circ}}}{\frac{\sin20^{\circ}}{\sin50^{\circ}}}=\frac{\sin50^{\circ}}{2\sin80^{\circ}\sin20^{\circ}}.$$ Thus, $$4\sin20^{\circ}\sin80^{\circ}\sin(80^{\circ}-x)=\sin(20^{\circ}+x)$$ or $$2(\cos60^{\circ}-\cos100^{\circ})\sin(80^{\circ}-x)=\sin(20^{\circ}+x)$$ or $$\sin(80^{\circ}-x)+\sin(160^{\circ}-x)-\sin{x}=\sin(20^{\circ}+x)$$ or $$\sin(80^{\circ}-x)=\sin{x},$$ which gives $$x=40^{\circ}.$$
In this picture, there are some things that have to be further explained:
First of all, we take the symmetry of $\Delta BEC$. Then we draw a line from $B$ to $B'$ in order to have an equilateral triangle $\Delta BEB'$. Then we have the equality $|EB| = |BB'| = |EB'| = |AB|$. Here, since $\angle EBB' = 60^\circ$ and $|AB| = |BB'|$, we can conclude that $A$, $F$, $D$ and $B'$ are linear and $\angle AB'B = 20^\circ$. Then by noticing the fact that $EO$ is median of the equilateral triangle, it is median of $\Delta BDB'$ as well, which implies that $\angle DBB' = 20^\circ$. Therefore $x = 40^\circ$.