Yet another geometry puzzle

 In triangle $ABC$, $\angle B=60^\circ$ and $\angle C=40^\circ$. Extend $A$ away from $C$ to $D$ such that $AD=BC$. What is $\angle ADB$?

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Solution:

Let point $E$ lie on segment $BC$ such that $AEB$ is an equilateral triangle. Let point $F$ lie on segment $AD$ such that $AF=AB$. Since SAS condition holds, $FBE$ is congruent to $DAB$, and $\angle ADB=30^\circ$ follows almost immediately.