2008 AMC 10B Problem 24

Let $M$ be the point with $\triangle BMC$ equilateral, on the same side of $BC$ as $A.$ Then $\angle ABM=70^\circ-60^\circ=10^\circ$ and $\angle MCD=170^\circ-60^\circ=110^\circ.$ Since $AB=BM$ and $MC=CD,$ triangles $ABM$ and $MCD$ are isosceles, giving $\angle AMB=85^\circ$ and $\angle CMD=35^\circ.$ Then $\angle AMD=360^\circ-85^\circ-60^\circ$ $-35^\circ$ $=180^\circ,$ so $M$ lies on $\overline{AD}$ and $\angle BAD=\angle BAM=85^\circ.$ Thus, the correct answer is C.