$\angle ACD$ is an exterior angle of $\Delta ABC$. If $\angle A = 70^{\circ}$ and $\angle B = 40^{\circ}$ in $\Delta ABC$,then find $\angle ACD$. (in $^{\circ}$)

Ray $BD$ is the bisector of $\angle ABC$ and ray $BE$ is the bisector of $\angle DBC$. If $\angle EBC = 19^{\circ},$ then find $\angle DBC, \angle ABC$ and $\angle ABE$.

$\angle ACD$ is an exterior angle of $\Delta ABC$. The bisector of $\angle ABC$ and exterior $\angle ACD$ intersect each other at point $E$. Prove that,$\angle BEC = \frac{1}{2} \angle BAC$.

In the given figure,$AB \parallel DE$ and $BC \parallel EF$. Prove that $\angle ABC + \angle DEF = 180^{\circ}$.

For what value of $x+y$ in the figure will $ABC$ be a straight line? Justify your answer.

Let $OA, OB, OC$ and $OD$ be rays in the anticlockwise direction such that $\angle AOB = 100^{\circ}, \angle COD = 100^{\circ}, \angle BOC = 82^{\circ}$ and $\angle AOD = 78^{\circ}$. Is it true to say that $AOC$ and $BOD$ are lines?