# Find $x$ in this $80^\circ$-$80^\circ$-$20^\circ$ triangle ($60^\circ$-$70^\circ$ variant) [duplicate]

Refer to the diagram and find x in degrees.

My method is to let AB=1, and express AD and AE in terms of AB using sine formula. Then find DE using cosine formula. After that use cosine formula to find cos x. Finally x=20 degrees. However this method relies on the calculator. Is there any PURELY geometric method to solve this problem, such as adding a line which is parallel to line AB, so that it can be PROVEN that x is 20 degrees, without the use of a calculator?

First off, we can find $A\hat FB=180-(70+60)=50^\circ$ because angles in a triangle must sum to $180^\circ$

Now we can use the vertially opposite rule to say that $D\hat FE$ is also $50^\circ$

Now, we can say that $A\hat FD=B\hat FE = (360-2\times 50) \div 2=130^\circ$

Next we can find $F\hat EB=180-(20+130)=30^\circ$ using angles in a triangle rule again

Now, we find $A\hat CB = 180- ((70+10)+(60+20))=20^\circ$

Next we can say that $C\hat EA = 180 - 30 = 150^\circ$ using the fact that angles on a line sum to $180^\circ$

This is the point at which I get slightly stuck, so I'll leave this here and see if you can get any further:

Note: All of the above assumes that $AE$ and $BD$ are straight lines