In the given figure, AB, BC, CD and DA are tangents to the circle with centre O forming a quadrilateral ABCD . Show…

Join $OP, OQ, OR$ and $OS$
$\triangle POB \cong \triangle QOB$
$\Rightarrow \angle 1 = \angle 2$
Similarly $\angle 3 = \angle 4, \angle 5 = \angle 6, \angle 7 = \angle 8$
Now, $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$
$\Rightarrow 2(\angle 1 + \angle 8 + \angle 4 + \angle 5) = 360^\circ$
$\therefore \angle AOB + \angle COD = 180^\circ$