Question

Prove that opposite sides of a quadrilatral circumscribing a ciecle subtend supplementary angle at the centre of the circle.

Answer 1

Given: Let ABCD be the quadrilateral circumscribing the circle with centre O.

ABCD touches the circle at points P, Q, R and S.

To prove: Opposite sides subtends supplementary angles at centre.

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

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

Construction: Join OP, OQ, OR and OS.

Proof: Let us rename the angles

In $\Delta AOP$ and $\Delta AOS$,

AP$=$AS

(Length of tangents drawn from external point to a circle are equal)

AO$=$AO (common)

OP$=$OS (both radius)

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

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

i.e., $\angle 1=\angle 8$ …………….$\left(1\right)$

Similarly, we can prove

$\angle 2=\angle 3$ ……………….$\left(2\right)$

$\angle 5=\angle 4$ ………………..$\left(3\right)$

$\angle 6=\angle 7$ ……………….$\left(4\right)$

Now,

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

(sum of angles round a point is ${360}^o$)

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

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

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

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

Hence both angles are supplementary.

similarly, we can prove

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

Hence proved.