To find angle BGE from the diagram above, a proof was proposed:
"Angle BCF equals 100 degrees (external to triangle ACB); ADEC and DBFC are cyclic quadrilaterals -> angle ADE equals 100 degrees; angle DEB equals 70 degrees. Let´s call angle CBG alpha. Then, angle DCA equals 30+alpha and angle DCB equals 50-alpha (30+alpha+100+DCB=180 -> DCB=50-alpha). But because DECA is cyclic, angle DAE equals angle DCE.
Then angle BEG equals 80-alpha (external to triangle ABE). Now, we have:
80-alpha+alpha+BGE=180 -> BGE= 100 degrees."
Regarding the proof, I found the same thing up until the alpha part, which I found very confusing (I do not know why some angles equal what the proof has stated them to be). Hence, I would really appreciate if someone could find a proof excluding alpha and just involving theorems.
This is an alternate proof I had written up, but I'm still stuck on what to do next:
Help would really be appreciated. Thanks :)
Here is the picture with hints.
Added:
All angles marked in red are equal (somehow). [By considering the large circle, the leftmost red is equal to rightmost red. By considering the small circle, the leftmost red is equal to the middle red. Note that their actual values need not be found because they will be equal definitely.]
From the above, we can say the bottom right quadrilateral is cyclic because of "exterior angle equals to the interior opposite angle". Since that quadrilateral is cyclic now, we can say that the green marked angle is $100^0$.