In the given figure,$POQ$ is a straight line. The value of $x$ is: (in $^{\circ}$)

$\angle ACD$ is an exterior angle of $\Delta ABC$ and the bisector of $\angle A$ intersects $BC$ at $E$. Prove that $\angle ABC + \angle ACD = 2 \angle AEC$.

$\angle A$ and $\angle B$ are complementary angles. If $\angle A = \angle B - 20^{\circ}$,then find $\angle A$ and $\angle B$.

An exterior angle of a triangle is $105^{\circ}$ and its two interior opposite angles are equal. Each of these equal angles is

Can a triangle have all angles less than $60^{\circ}$? Give reason for your answer.

$A$ transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.