Problem: Consider the triangle $ABC$, $\angle A = \angle B = 40^\circ$. The point $D$ on $AC$ is such that $\angle ABD=10^\circ$. Consider the point $E$ on $AB$ such that $BC=EB$. Prove that $DE \parallel BC$.
Context: This is the reformulation (the conclusion is simplified) of an exercise from 2022 local competition. The official solution has -- in my opinion -- a big gap. They say:
Let $A'$ be symmetric to $A$ wrt $CE$. Then $AA'C$ is equilateral. From the sum of measures of angles in the triangle $AA'D$ we get that $\angle ADA'=100^\circ$. Then $DE \parallel BC$. $\square$
The problem is that they took for granted that $E$ lies on $A'D$. This is equivalent to the conclusion and I don't understand how considering $A'$ can help.
My tries: Conclusion $\iff \angle AED = 40^\circ\iff \angle EDA = 100^\circ \iff AD=ED \iff E\in A'D$.
Let $F$ be the intersection of $CA'$ and $AB$.
The goal is to show that $DC = CF$.
From there, $F$ is the reflection of $D$ along $CE$ and $AEF$ is a straight line implies that it's reflection $A'ED$ is also a straight line (which is what OP wanted to show).
Let $BC = 1$.
Apply sine rule in triangle $BCD$: $DC / \sin 30^\circ = BC / \sin 50^\circ$, so $ DC = \frac{1}{2 \sin 50^\circ}$.
Apply sine rule in triangle $BCF$: $CF/ \sin 40^\circ = BC / \sin 100^\circ$, so $CF = \frac{ \sin 40^\circ } { \sin 100^\circ}$.
Since $ \frac{\sin 40 ^\circ} { \sin 100^\circ} = \frac{ \sin 40 ^\circ } { 2 \sin 50^\circ \cos 50^\circ } = \frac{ 1 } { 2 \sin 50 ^\circ}$, so $DC = CF$ as desired.
Notes