2015 AMC 12B Problem 13

Angles $BAC$ and $BDC$ subtend arc $BC,$ so $\angle BDC = 70^\circ.$ Then $\angle ADC = \angle ADB$ $+ \angle BDC = 110^\circ.$ Since $ABCD$ is cyclic, $\angle ABC = 180^\circ - 110^\circ = 70^\circ$ $= \angle BAC.$ Thus $\triangle ABC$ is isosceles with $AC = BC = 6.$ Thus, the correct answer is B.