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Find angle $ \angle AED $ in the following triangle.enter image description here

In the above triangle we have : $CA=CB ,CE=DB=BA ,\angle ACB =20^° , \angle CAB=\angle CBA=80^°$ now find $ \angle AED $.

I think if we draw line $AD$ we have $AB=BD$ then $\angle BAD= \angle BDA =50^°$ then in the triangle $ AED$ we have $\angle EAD=30^°$ but I can't find $ \angle AED $

amir bahadory
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Clathratus, how did you prove that AEDB is cyclic since you have used the external angle theorem of a cyclic quadrilateral when you say that angle CED is equal to angle DBA ?

Anonymous
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