$A$ 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 $m \angle AOD + m \angle BOC = 180^{\circ}$.

(N/A) circle with centre $O$ touches the sides $\overline{AB}$,$\overline{BC}$,$\overline{CD}$ and $\overline{DA}$ of quadrilateral $ABCD$ at $P, Q, R$ and $S$ respectively.
To prove: $m \angle AOB + m \angle COD = 180^{\circ}$ and $m \angle AOD + m \angle BOC = 180^{\circ}$.
Proof: Draw $\overline{OP}, \overline{OQ}, \overline{OR}$ and $\overline{OS}$.
Since tangents from an external point to a circle are equal,$\overline{AS} \cong \overline{AP}$.
In $\Delta ASO$ and $\Delta APO$:
$\overline{AS} \cong \overline{AP}$ (Tangents from point $A$)
$\overline{OS} \cong \overline{OP}$ (Radii of the same circle)
$\overline{AO} \cong \overline{AO}$ (Common side)
Therefore,$\Delta ASO \cong \Delta APO$ ($SSS$ congruence rule).
This implies $m \angle OAS = m \angle OAP$,so $m \angle OAB = \frac{1}{2} m \angle DAB$.
Similarly,we can show:
$m \angle OBA = \frac{1}{2} m \angle ABC$
$m \angle OCD = \frac{1}{2} m \angle BCD$
$m \angle ODC = \frac{1}{2} m \angle CDA$
In $\Delta AOB$,$m \angle AOB = 180^{\circ} - (m \angle OAB + m \angle OBA) = 180^{\circ} - \frac{1}{2}(m \angle DAB + m \angle ABC)$.
In $\Delta COD$,$m \angle COD = 180^{\circ} - (m \angle OCD + m \angle ODC) = 180^{\circ} - \frac{1}{2}(m \angle BCD + m \angle CDA)$.
Adding these:
$m \angle AOB + m \angle COD = 360^{\circ} - \frac{1}{2}(m \angle DAB + m \angle ABC + m \angle BCD + m \angle CDA)$.
Since the sum of angles in a quadrilateral is $360^{\circ}$:
$m \angle AOB + m \angle COD = 360^{\circ} - \frac{1}{2}(360^{\circ}) = 360^{\circ} - 180^{\circ} = 180^{\circ}$.
Similarly,it can be proved that $m \angle AOD + m \angle BOC = 180^{\circ}$.