Question

A circle touches the sides of a quadrilateral $ABCD$ at the points $P,Q,R$ and $S$ respectively. Show that the angles subtended at the center by by a pair of opposite sides are supplementary.

Answer 1

Give : Let $ABCD$ be the quadrilateral circumscribing the circle with centre $O$. $ABCD$ touches the circle at point $P,Q,R$ & $S$.

To Prove : opposite angle subtend supplementary angles at center

i.e. $\angle AOB+\angle COD={180}^o$

& $\angle AD+\angle BOC={100}^o$

Proof : In $\Delta AOP$ and $\Delta AOS$

$AP=AS$

$AO=AO$ (common)

$OP=OS$ (both radius)

$\Delta AOP=\Delta AOS$ (SSS congruence rule)

$\angle AOP=\angle AOS$ (CPCT)

i.e. $\angle 1=\angle 8$ ...(i)

Similiarly, we have

$\angle 2=\angle 3$ ...(ii)

$\angle 5=\angle 4$ ...(iii)

$\angle 6=\angle 7$ ...(iv)

Now

$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360$

$\angle 1+\angle 2+\angle 2+\angle 5+\angle 5+\angle 6+\angle 6+\angle 1=360$

$2[\angle 1+\angle 2+\angle 5+\angle 6]={360}^o$

$\angle 1+\angle 2+\angle 5+\angle 6={180}^o$

$(\angle 1+\angle 2)+(\angle 5+\angle 6)={180}^o$

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

Hence both angles are supplementary.

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

Hence proved.