Prove that in a cyclic quadrilateral opposite angles are supplementary.
Given: ABCD is a cyclic quadrilateral of a circle with centre O.
To prove:
∠ABC + ∠ADC = 180°
and ∠BAD + ∠BCD = 180°
Construction: A, O, and C, O are joined.
Proof:
The reflex angle ∠AOC at the centre and the angle ∠ABC on the circle are formed with the circular arc ADC.
∴ Reflex ∠AOC = 2∠ABC
∴ ∠ABC = 1\over 2 reflex ∠AOC ………(i)
Again, ∠AOC is the angle at the centre and ∠ADC is the angle on the circle formed with the circular arc ABC.
∴ ∠AOC = 2∠ADC
∴ ∠ADC = 1\over 2 ∠AOC ……….(ii)
From (i) and (ii), we get:
∠ABC + ∠ADC = 1\over 2 reflex ∠AOC + 1\over 2 ∠AOC
= 1\over 2 (reflex ∠AOC + ∠AOC)
= 1\over 2 × 4 right angles
= 180°
Similarly, by joining B, O, and D, O; it can be proved that,
∠BAD + ∠BCD = 180° (proved).
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