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In the figure, diagonals AC and BD of a parallelogram ABCD intersect each other at point O, such that $$\angle DAC = 45^\circ$$ and $$\angle AOB = 110^\circ$$. The measure of $$\angle DBC$$ is
Since $$\angle AOB$$ and $$\angle AOD$$ form a linear pair, $$\angle AOD = 70^\circ$$. In $$\triangle AOD$$, $$\angle ADO = 180^\circ - 45^\circ - 70^\circ = 65^\circ$$. As $$AD \parallel BC$$, $$\angle DBC = \angle ADB = 65^\circ$$ (alternate angles).
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