The diagonals $AC$ and $BD$ of a parallelogram $ABCD$ intersect each other at the point $O$. If $\angle DAC=32^{\circ }$ and $\angle AOB=70^{\circ }$, then $\angle DBC$ is equal to

Since $ABCD$ is a parallelogram, line $AD\parallel BC$, and diagonal $AC$ acts as a transversal line.Therefore, alternate interior angles are equal:
$$\angle BCA=\angle DAC=32^{\circ }$$
In $△OBC$, the exterior angle at vertex $O$ for this triangle is $\angle AOB=70^{\circ }$ (since $A,O,C$ lie on a straight line, $\angle BOC=180^{\circ }-70^{\circ }=110^{\circ }$).Using the exterior angle theorem for $△OBC$:
$$\angle AOB=\angle OBC+\angle BCA$$
$$70^{\circ }=\angle OBC+32^{\circ }$$
$$\angle OBC=70^{\circ }-32^{\circ }=38^{\circ }$$
Since $\angle OBC$ is the exact same angle as $\angle DBC$:
$$\angle DBC=38^{\circ }$$