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In the given figure, a circle is inscribed in a quadrilateral $ABCD$ which touches the sides $AB, BC, CD$ and $DA$ at $P, Q, R$ and $S$ respectively. Prove that $\angle AOB + \angle COD = 180^\circ$.
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Proving $\Delta OAP \cong \Delta OAS$ (by any congruency criterion) [$1$ mark]
$\Rightarrow \angle 1 = \angle 6$ (cpct)
Similarly $\angle 3 = \angle 5, \angle 4 = \angle 7$ and $\angle 2 = \angle 8$ [$1$ mark]
Also $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$ [$\frac{1}{2}$ mark]
$\therefore \angle 1 + \angle 2 + \angle 3 + \angle 4 = 180^\circ$
$\angle AOB + \angle COD = 180^\circ$ [$\frac{1}{2}$ mark]
$\Rightarrow \angle 1 = \angle 6$ (cpct)
Similarly $\angle 3 = \angle 5, \angle 4 = \angle 7$ and $\angle 2 = \angle 8$ [$1$ mark]
Also $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$ [$\frac{1}{2}$ mark]
$\therefore \angle 1 + \angle 2 + \angle 3 + \angle 4 = 180^\circ$
$\angle AOB + \angle COD = 180^\circ$ [$\frac{1}{2}$ mark]