(a) Steps: 2 Marks
      Result: 1 Mark
(b) Solution: 1 Mark
(a) Since $AB\parallel CD$
$\Rightarrow \angle BAC+\angle DCA={180}^{\circ }$  
[Sum of Co-interior angles is equal to ${180}^{\circ }$]
$\Rightarrow \frac{\angle BAC}{2}+\frac{\angle DCA}{2}=\frac{{180}^{\circ }}{2}$
$\Rightarrow \angle OAC+\angle OCA={90}^{\circ }$   [AD & BC are angle bisectors]
In $\Delta AOC,$
$\angle AOC+\angle OAC+\angle OCA={180}^{\circ }$
 [sum of  all the angles in triangle is ${180}^{\circ }$]
$\Rightarrow \angle AOC={180}^{\circ }-{90}^{\circ }={90}^{\circ }$
$\angle AOC=\angle BOD$  [Vertically Opposite Angles]
$\Rightarrow \angle BOD={90}^{\circ }$
(b)  The value of an exterior angle of a triangle is equal to the sum of the opposite interior angles.

As the exterior angle value is given to be 120${}^{\circ }$, the sum of the two interior angles is also 120${}^{\circ }$.
Thus, (x+y) = 120${}^{\circ }$