Theorem 10.11
The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
Given : ABCD is a cyclic quadrilateral.
of a circle with centre at O
To Prove : ∠ BAD + ∠ BCD = 180°
∠ ABC + ∠ ADC = 180°
Proof:
Chord AB
Angles in same
segment are equal.
∠5 = ∠8
Chord BC
Angles in same
segment are equal.
∠1 = ∠6
Chord CD
Angles in same
segment are equal.
∠2 = ∠4
Chord AD
Angles in same
segment are equal.
∠7 = ∠3
∠1 + ∠2 + ∠3 + ∠4 + ∠7 + ∠8 + ∠5 + ∠6 = 360°
(∠1 + ∠2 + ∠7 + ∠8) + (∠3 + ∠4 + ∠5 + ∠6) = 360°
∴ (∠1 + ∠2 + ∠7 + ∠8) + (∠7 + ∠2 + ∠8 + ∠1) = 360°
⇒ 2 (∠1 + ∠2 + ∠7 + ∠8) = 360°
∠1 + ∠2 + ∠7 + ∠8 = 180°
(∠1 + ∠2) + (∠7 + ∠8) = 180°
∠BAD + ∠BCD = 180°
Similarly,
∠ABC + ∠ADC = 180°
Hence, Proved.
From (1), (2) , (3), (4)
∠ 3 = ∠ 7
∠ 4 = ∠ 2
∠ 6 = ∠ 1
∠ 5 = ∠ 8

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.