Question

Prove that, the sum of either pair of opposite angles of a cyclic quadrilateral is 180.

Answer 1

Given : ABCD is a cyclic quadrilateral of a circle with centre at O.

To prove : $\angle BAD+\angle BCD={180}^o$

$\angle ABC+\angle ADC={180}^o$

Chord $AB$

$\angle 5=\angle 8.....\left(1\right)$ [Angle in same segment are equal]

Chord $BC$

$\angle 1=\angle 6.....\left(2\right)$ [Angle in same segment are equal]

Chord $CD$

$\angle 2=\angle 4.....\left(3\right)$ [Angle in same segment are equal]

Chord $AD$

$\angle 7=\angle 3.....\left(4\right)$ [Angle in same segment are equal]

By angle sum property of qudrilateral

$\angle A+\angle B+\angle C+\angle D={360}^o$

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

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

$(\angle 1+\angle 2+\angle 7+\angle 8)+(\angle 7+\angle 2+\angle 8+\angle 1)={360}^o$ [From (1), (2) ,(3) and (4)]

$2(\angle 1+\angle 2+\angle 7+\angle 8)={360}^o$

$\angle 1+\angle 2+\angle 7+\angle 8={180}^o$

$(\angle 1+\angle 2)+(\angle 7+\angle 8)={180}^o$

$\angle BAD+\angle BCD={180}^o$

Similarly,

$\angle ABC+\angle ADC={180}^o$

Hence proved.