jmc

Since $\overline{BC}\cong \overline{DC}$, that means $\angle DBC\cong \angle BDC$ and $$m\angle DBC=m\angle BDC=70^{\circ }.$$ We see that $\angle BDC$ and $\angle ADC$ must add up to $180^{\circ }$, so $m\angle ADC=180-70=110^{\circ }$. Triangle $ACD$ is an isosceles triangle, so the base angles must be equal. If the base angles each have a measure of $x^{\circ }$, then $m\angle ADC+2x=180^{\circ }.$ This gives us $$110+2x=180,$$ so $2x=70$ and $x=35.$ Since $\angle BAC$ is one of the base angles, it has a measure of $\boxed{35^{\circ }}$.