$A$ circle touches all the sides of a quadrilateral $ABCD$. Then, the quadrilateral $ABCD$ is a:

$\overline{AB}$ is a chord of a circle with centre $O$. Line $l$ touches the circle at $B$. The foot of the perpendicular from $A$ to $l$ is $D$. Prove that $\angle BAO \cong \angle BAD$.

$\overline{AB}$ is a chord of $\odot(O, 13)$ such that $AB = 24$. Tangents at $A$ and $B$ to the circle intersect at $P$. Find $PA$.

Point $A$ lies in the exterior of $\odot(O, 8)$. $A$ line through $A$ touches the circle at $B$. If $AB = 15$,then find $OA$.

$\odot(O, 41)$ and $\odot(O, 9)$ are concentric circles. The chord $\overline{AB}$ of $\odot(O, 41)$ touches $\odot(O, 9)$ at point $M$. Then $AB = \ldots$

In the figure,if $O$ is the centre of a circle,$PQ$ is a chord,and the tangent $PR$ at $P$ makes an angle of $50^{\circ}$ with $PQ$,then $\angle POQ$ is equal to: (in $^{\circ}$)