The diagonals AC and BD of a parallelogram ABCD intersect each other at the…

Concept: In a parallelogram, opposite sides are parallel, so a diagonal acting as a transversal creates equal alternate interior angles between them. The exterior-angle theorem also states that an exterior angle of a triangle equals the sum of the two interior angles not adjacent to it, and angles forming a linear pair on a straight line sum to 180°.

1. O lies on diagonal AC, so ∠OAD is the same angle as the given ∠DAC = 32°.

2. ∠AOB and ∠AOD lie along the straight diagonal BD, so they form a linear pair: ∠AOD = 180° − 70° = 110°.

3. In triangle AOD, the angles sum to 180°: ∠ADO = 180° − 32° − 110° = 38°.

4. Since AD ∥ BC, diagonal BD is a transversal between them, so the alternate interior angles ∠ADB (= ∠ADO) and ∠DBC are equal.

5. Therefore, ∠DBC = 38°.

Cross-check: ∠AOB is also the exterior angle of triangle AOD at O (it is supplementary to the interior ∠AOD), so by the exterior-angle theorem ∠AOB = ∠OAD + ∠ODA, i.e. 70° = 32° + ∠ADO, giving ∠ADO = 38° again — confirming ∠DBC = 38° by an independent route.