geometry
We are looking for angles x and y.
I have found the values of the following angles:
BEA = 74,
BDA = 64,
ACD = 68,
ECD = 112,
plus the relationship $x+y = 68$.
All other angles equations, from triangles or the sum of angles in the quadrilateral (360) end up in the same equation!
I have found through Geogebra that $x=18$ and $y=50$ but I can't figure out a second relationship to determine them geometrically!
Does anyone have any ideas?
Thank you!
I have four different and one general solution to this problem, I hope it will help you.
$\def\C{{\cal C}}$ Let $\C_A$ and $\C_D$ be the circumscribed circles of $\triangle ABC$ and $\triangle DBC$, respectively.
Assume these circles do not coincide: $\C_A\ne\C_D$.
Since the circles intersect at points $B$ and $C$ one of them is inside the other in the upper half-plane (the half-plane created by the line $BC$, where the points $A$ and $D$ are situated). Without loss of generality we may assume it is the circle $\C_D$. This means that $D$ is an inner point of the circle $\C_A$. Since the quadrilateral is assumed to be convex the points $B$ and $D$ are in different half-planes created by the line $AC$ (see figure).
Continue the lines $CD$ and $AD$ till intersection with the circle $\C_A$. Let $F=(CD)\cap\C_A$, $G=(AD)\cap\C_A$. Observe that $G$ is an inner point of the arc $FC$.
In view of $\angle ABC=\angle BCF$ and $\angle BAC=\angle BFC$ we have: $$ \triangle ABC\cong \triangle FCB\implies \angle BCA=\angle CBF. $$
Hence: $$ \angle DAE=\angle GAC=\frac12 \overset{\mmlToken{mo}{⏜}}{GC} <\frac12 \overset{\mmlToken{mo}{⏜}}{FC} =\frac12\overset{\mmlToken{mo}{⏜}}{AB}=\angle AGB<\angle ADB=\angle ADE, $$ which contradicts to the condition that $AED$ is isosceles triangle.
Thus, the assumption $\C_A\ne \C_D$ was false, and the quadrilateral $ABCD$ is in fact cyclic. The conclusion $AD\parallel BC$ follows immediately.
Best Answer
If you are looking for the proof by elementary geometry, please see Hiroshi Saito(斉藤浩)'s great work about generalized Langley's problem ( https://www.gensu.co.jp/saito/challenge/pdf/3circumcenter_d20180609.pdf ). He introduced the amazing skill named "3 circumcenter method" invented by Ms. aerile_re(pen name), and you can find the solution of this problem in the article (Q1).