This is the original drawn of this question:
You will easily notice that AFE angle is 20. And AFE and FCA triangles are isosceles triangle:
Moreover, if you draw a line through point E as 20 degrees as seen below, we have another isosceles triangle FHE and equillateral triangle AIE:
You will probably already notice that FCA and FHE triangles are the same triangles but reflections. Therefore ACE angle and EHA angles are equal to each other and if we join the point H and C, it has to be parallel to AE line.
As a result, CHI triangle has to be equillateral triangle because HIC angle is 60 degrees and HI and CI are equal to each other
On the other hand, the red lines are equal to each other because ADE and AED angles are equal to each other. Moreover, JDE triangle becomes isosceles since JDE and JED angles are 30 degree:
Lastly, if we connect point A to point K as shown below, we will have another triangle KCA. We already know that KHA angle is 80 degrees from previous drawings, and KDH and ADI are 80 degrees as well. So as a result, |KH|=|KD|, and |AD|=|AI|, ann we know that CHI triangle is equillateral, we can conclude that |KH|=|KD|=|AD|=|AI| and C ray passing just between AK line is bisector and the result becomes:
Note: I will fix mathjax part later and explain it more. (out of time right now :))