Another geometry puzzle

 $ABC$ is a isosceles triangle with $AC=BC$ and $\angle C=20^\circ$. Point $D$ lies inside segment $BC$ such that $DC=AB$. What is $\angle ADB$?

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Solution:

Let point $E$ lie on the same side of segment $AC$ as $B$ such that $AEC$ is a equilateral triangle. Due to SAS condition, $ABE$ is congruent to $CDA$. It then follows almost immediately that $\angle ADB=30^\circ$.