The diagonals AC and BD of a parallelogram
Question:

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point 0. If ∠DAC = 32° and ∠AOB = 70°, then ∠DBC is equal to

(a) 24°

(b) 86°

(c) 38°

(d) 32°

Solution:

(c) Given, $\angle A O B=70^{\circ}$ and $\angle D A C=32^{\circ}$

$\therefore \quad \angle A C B=32^{\circ} \quad[A D \| B C$ and $A C$ is transversal $]$

Now, $\quad \angle A O B+\angle B O C=180^{\circ} \quad$ [linear pair axiom]

$\Rightarrow \quad \angle B O C=180^{\circ}-\angle A O B=180^{\circ}-70^{\circ}=110^{\circ}$

Now, in $\triangle B O C$, we have

$\angle B O C+\angle B C O+\angle O B C=180^{\circ}$ [ by angle surn property of a triangle]

$\Rightarrow \quad 110^{\circ}+32^{\circ}+\angle O B C=180^{\circ} \quad\left[\because \angle B C O=\angle A C B=32^{\circ}\right]$

$\Rightarrow \quad \angle O B C=180^{\circ}-\left(110^{\circ}+32^{\circ}\right)=38^{\circ}$

$\therefore \quad \angle D B C=\angle O B C=38^{\circ}$