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

If from an external point $B$ of a circle with centre $O$,two tangents $BC$ and $BD$ are drawn such that $\angle DBC = 120^{\circ}$,prove that $BC + BD = BO$,i.e.,$BO = 2BC$.

Write 'True' or 'False' and give reasons for your answer.
$AB$ is a diameter of a circle and $AC$ is its chord such that $\angle BAC = 30^{\circ}$. If the tangent at $C$ intersects $AB$ extended at $D$,then $BC = BD$.

If $P$ is a point in the exterior of the circle,then maximum $\ldots \ldots \ldots \ldots$ tangents can be drawn to a circle from $P.$

$A$ circle with centre $O$ touches sides $\overline{AB}$,$\overline{BC}$,$\overline{CD}$ and $\overline{DA}$ of quadrilateral $ABCD$ at points $P, Q, R$ and $S$ respectively. Prove that $m \angle AOB + m \angle COD = 180^{\circ}$ and $m \angle AOD + m \angle BOC = 180^{\circ}$.

$\overleftrightarrow{PA}$ and $\overleftrightarrow{PB}$ are tangents to the circle $\odot(O, r)$ at $A$ and $B$ respectively. If $m\angle AOB = 100^\circ$,then $m\angle OPB = \dots$ (in $^\circ$)