Let ABD = CBD = a. Extend ray BA to X and BC to Y. Simple angle chasing gives us DAX = DAC = 66, or AD is the exterior angle bisector of BAC. So, D is the B-excenter of triangle ABC. Thus, DCA = DCY = b (say). Now let us perform some trivial angle chasing. Using the exterior angle sum property, we get b = a + BDC and 66 = a + ADB. So, b + 66 = 2a + ADC, or, 180 - ADC = 2a + ADC, or, a + ADC = 90. Hence, we get BDC = ADC - ADB = 90 - a - ADB = 90 - 66 = 24.

Excellent observation, Alex; to complete, if $I$ was the incenter of $\triangle ABC$, then $AICD$ is cyclic, and $\angle BDC=\angle IAC=24^\circ$, done. Best regards, sunken rock

Oh...that was even better