Question

A circle touch the sides of a quadrilateral $ABCD$ at $P,Q,R,S$ respectively Show that the angle subtended at the centre by a pair of opposite sides are supplementary.

Answer 1

A circle the centre $O$ touches the sides $AB$, $BC$, $CD$ and $DA$ of a quadrilateral $ABCD$ at the points $P,Q,R$ and $S$ respectively.

To prove: $\angle AOB+\angle COD={180}^o$
and, $\angle AOD+\angle BOC={180}^o$

Construction: Join $OP,OQ,OR$ and $OS$

Proof:

Since the two tangents drawn from an external point to a circle subtend equal angles at the centre.

$\therefore \angle 1=\angle 2,\angle 3=\angle 4,\angle 5=\angle 6$ and $\angle 7=\angle 8$

Now, $\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8={360}^o$

$\Rightarrow 2(\angle 2+\angle 3+\angle 6+\angle 7)={360}^o$ and

$2(\angle 1+\angle 8+\angle 4+\angle 5)={360}^o$

$(\angle 2+\angle 3+)+(\angle 6+\angle 7)={180}^o$ and $(\angle 1+\angle 8)+(\angle 4+\angle 5)={180}^o$

[$\because \angle 2+\angle 3=\angle AOB,\angle 6+\angle 7=\angle COD,\angle 1+\angle 8=\angle AOD$ \ and \ $\angle 4+\angle 5=\angle BOC$]

$\Rightarrow \angle AOB+\angle COD={180}^o$

$\Rightarrow \angle AOD+\angle BOC={180}^o$