If an angle is $20^{\circ}$ more than its complementary angle,then the measure of that angle is $\ldots \ldots \ldots$ (in $^{\circ}$)

In $\Delta ABC$,the bisectors of $\angle B$ and $\angle C$ intersect each other at $I$. Prove that $\angle BIC = 90^{\circ} + \frac{1}{2} \angle A$.

Two lines $l$ and $m$ are perpendicular to the same line $n$. Are $l$ and $m$ perpendicular to each other? Give reason for your answer.

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 $\angle ABE$.

In the given figure,if $AB \parallel CD$,$\angle PCD = 130^{\circ}$ and $\angle PBA = 140^{\circ}$,then find $\angle BPC$. (in $^{\circ}$)

$\angle A = \angle B$. If $\angle A$ and $\angle B$ are supplementary angles,then $\angle A = \ldots$ (in $^o$)