Let $BA= BC$ and $BD = AC$. Find angle $BDC$. (See image) 

Let $BA= BC$ and $BD = AC$. Find angle $BDC$.

This is probability a very elementary question, and I'm thinking of using sine rule to brute force it. However, I seem to feel like there should be an elegant solution to this that involves shifting the triangles around (perhaps side BD to AC). Is there any simple way of doing this?
 A: 
A very famous problem. This is "the way" to solve it in my opinion.
Construct $B'$ so that $\triangle BDB'$ is congruent to $\triangle ACB$. Notice how $\angle B'BC$ is $60^{\circ}$ and the pink triangle is equilateral and the three pink edges are of the same length.
A: Let in $\Delta ACB'$, where $B$ and $B'$ are placed in the same half-plane respect to $AC$,
we have $AB'=CB'$, $E\in AB$, $D'\in B'E$ and $F\in CB'$ such that $$AC=CE=EF=FD'=D'B'.$$
Thus, easy to show that $$\measuredangle AB'C=20^{\circ}$$ and $$\measuredangle B'AC=\measuredangle B'CA=80^{\circ},$$
which gives $B'\equiv B$, $D'\equiv D$ and $$\measuredangle CDB=\measuredangle A+(\measuredangle ACB-\measuredangle FCD)=80^{\circ}+(80^{\circ}-10^{\circ})=150^{\circ}.$$
A: Let $X$ be such that $ACBX$ is an isosceles trapezoid with bases $BC$, $AX$. Clearly $\angle DBX = 80^\circ - 20^\circ = 60^\circ$ and $XB=AC=BD$, hence $BDX$ is equilateral. Therefore $DB=DX$. Also $XC=AB=CB$. Hence $CD$ is perpendicular bisector of $BX$. Since $\angle XDB=60^\circ$, it readily follows that $\angle BDC=150^\circ$.
