$\triangle ABC$ is clearly isosceles with base angle $\angle BAC=\angle BCA=40^{\circ}.$ Let the perpendicular from $B$ to $AC$ cut $AD$ at $E.$ $\triangle AEC$ is isosceles with apex angle $\angle AEC=120^{\circ}$ $\Longrightarrow$ $\angle CEB=\angle DEC=60^{\circ},$ i.e. $ED$ bisects $\angle BEC$ externally. Since $\angle ABE=50^{\circ},$ then $\angle EBD=75^{\circ}-50^{\circ}=25^{\circ}$ $\Longrightarrow$ $BD$ bisects $\angle EBC$ internally. Thus, $D$ is the B-excenter of $\triangle BEC$ $\Longrightarrow$ $\angle BDC=\frac{_1}{^2}\angle BEC=30^{\circ}.$

Here is another solution: Let $E$ be such point that $\Delta BCE$ is equilateral. Since $BA=BC=BE$ triangle $ABE$ is isosceles with $\angle BAE = 70^{\circ}$. This means that $E$ lies on side $AD$. But now $\Delta BDE$ is isosceles ($\angle EBD =\angle BDE = 35^{\circ}$) so $E$ is center of circle circumscribed for $\Delta BCD$. Thus $\angle BDC = {1\over 2}\angle BEC = 30^{\circ}$.

Take $F$ so that $ACF$ is an equilateral triangle, $B$ and $F$ on one side and the other of $AC$. Obviously, $D$ is on the perpendicular bisector of $AF$ and, as Luis proved, $BD$ is angle bisector of $\angle CBF$, that means $BCDF$ is cyclic, hence $\angle BFC=\angle BDC=30^\circ$. Best regards sunken rock