Question

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer 1

Step 1: Draw appropriate diagram and use the concept of congruence:

As we know that by theorem the tangents from the external point are equal

In $△OAP,△OAS$

So, $AP=AS$

$\Rightarrow OA=OA$ (It is the common side)

$\Rightarrow OP=OS$ (They are the radii of the circle)

So, by $SSScongruency△OAP\cong △OAS$

Thus,$\angle POA=\angle AOS$

This shows that$\angle 7=\angle 8$

In this same manner, the angles which are equal are

$$\begin{gathered} \Rightarrow \angle 4=\angle 3 \\ \Rightarrow \angle 2=\angle 1 \\ \Rightarrow \angle 6=\angle 5 \end{gathered}$$

Step 2: Use the concept of complete angle

On adding these angles we get,

$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360°\left[\mathrm{Complete}\mathrm{angle}\right]$

On rearranging,

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

$\Rightarrow$ $2\angle 1+2\angle 8+2\angle 5+2\angle 4=360°$

On taking out $2$ as common we get,

$\Rightarrow$ $(\angle 1+\angle 8)+(\angle 5+\angle 4)=180°$

Thus,$\angle BOC+\angle DOA=180°$

Similarly,$\angle AOB+\angle DOC=180°$

Hence, the opposite sides of any quadrilateral which is circumscribing a given circle will subtend supplementary angles at the center of the circle.