Suppose $\angle BAC = 60^\circ$ and $\angle ABC = 20^\circ$. A point $E$ inside $ABC$ satisfies $\angle EAB=20^\circ$ and $\angle ECB=30^\circ$.

Problem statement:

In a triangle $ABC$ with angles $\angle BAC = 60^\circ$ and $\angle ABC = 20^\circ$, a point $E$ inside the triangle is given such that $\angle EAB = 20^\circ$ and $\angle ECB = 30^\circ$. Prove that $E$ lies on an angle bisector of $\angle ABC$.

I drew a picture in Geogebra for this problem and this is what I did so far:

All angles drawn were found out by using the fact that the sum of angles in a triangle is $180^\circ$. I also noticed that $AD=BD$, $AC=CD=AE$ by using isosceles triangles.

I tried proving that $E$ lies on an angle bisector of $\angle ABC$ by proving that $GE=EI$ and so I drew perpendicular bisectors from $E$ to the sides $AB$ and $BC$, and I noticed two pairs of similar triangles: $\triangle AHE \sim \triangle AEF$ and $\triangle CJE \sim \triangle CED$, but I'm not sure if this is useful in any way.

I'm stuck since and I don't know how to continue on from this. I'm not sure if this is even the right approach to the problem. Is there a way of approaching this problem that I missed?

Suppose $\angle BAC = 60^\circ$ and $\angle ABC = 20^\circ$. A point $E$ inside $ABC$ satisfies $\angle EAB=20^\circ$ and $\angle ECB=30^\circ$.

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$\triangle ABC$ with a point $D$ inside has $\angle BAD=114^\circ$, $\angle DAC=6^\circ$, $\angle ACD=12^\circ$, and $\angle DCB=18^\circ$.

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