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

1. In a parallelogram $ABCD$, opposite sides $AD$ and $BC$ are parallel ($AD\parallel BC$).2. The diagonal $AC$ acts as a transversal line cutting across these parallel lines. Therefore, alternate interior angles are equal:
$$\angle ACB=\angle DAC=32^{\circ }$$
Consider $△OBC$. The angle $\angle AOB$ and $\angle BOC$ form a linear pair along the diagonal line segment $AC$:
$$\angle BOC=180^{\circ }-\angle AOB=180^{\circ }-70^{\circ }=110^{\circ }$$
Use the angle sum property for $△OBC$:
$$\angle DBC+\angle ACB+\angle BOC=180^{\circ }$$
$$\angle DBC+32^{\circ }+110^{\circ }=180^{\circ }$$
$$\angle DBC+142^{\circ }=180^{\circ }⟹\angle DBC=38^{\circ }$$