A geometry puzzle

 $ABC$ is a triangle where $\angle BAC=100^\circ$. Also $BA=AC$. Extend $C$ away from $A$ to $D$ such that $AD=BC$. What is $\angle BDA$?

This got me :(

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Let $BEA$ be an equilateral triangle such that $E,C,D$ lie on the same side of line $AB$. So $\angle EAC=40^\circ$, and triangle $ABC$ is congruent to triangle $EAD$ because SAS condition holds. Now $\angle BDA=30^\circ$ follows almost immediately.