$\overline{PA}$ and $\overline{PB}$ are the tangents to $\odot(O, r)$ drawn from a point $P$ outside a circle. If $m \angle APB = 70^\circ$,then $m \angle POB = \dots$ (in $^\circ$)

Let $s$ denote the semi-perimeter of a triangle $ABC$ in which $BC = a, CA = b, AB = c$. If a circle touches the sides $BC, CA, AB$ at $D, E, F$ respectively,prove that $BD = s - b$.

In the figure,the pair of tangents $AP$ and $AQ$ drawn from an external point $A$ to a circle with centre $O$ are perpendicular to each other and the length of each tangent is $5 \, cm$. Then the radius of the circle is (in $cm$):

$\overline{ PA }$ and $\overline{ PB }$ are tangents to the $\odot( O , r)$ drawn from a point $P$ outside the circle,then ..... .

If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at an angle of $80^{\circ}$,then $\angle POA$ is equal to ......$^{\circ}$

$\overline{ PA }$ and $\overline{ PB }$ are the tangents to $\odot( O , r)$ drawn from a point $P$ outside a circle. If $m \angle APB = 65^{\circ}$,then $m \angle AOB = \ldots \ldots \ldots . .$ (in $^{\circ}$)