NCERT Solutions
Class 9 Maths
Circles
Ex.10.6 Q.4
Let the vertex of an angle ABC be located outside a circle and
let the sides of the angle intersect equal chords AD and CE with the circle.
Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Given: Two equal chords AD and CE of a circle with centre O.
When meet at B when produced.
To Prove: ∠ABC = (∠AOC – ∠DOE) ÷ 2
Proof: Let ∠AOC = x, ∠DOE = y, ∠AOD = z
∠EOC = z
[Equal chords subtend equal angles at the centre]
=> x + y + 2z = 360
[Angle at a point] ………... (1)
OA = OD
=> ∠OAD = ∠ODA
Now in DOAD, we have
∠OAD + ∠ODA + z = 1800
=> 2∠OAD = 1800 – z
[Since ∠OAD = ∠OBA]
=> ∠OAD = 900 – 
…………... (2)
Similarly, ∠OCE = 90° – …...... (3)
=> ∠ODB = ∠OAD + ∠ODA
[Exterior angle property]
=> ∠OEB = 900 – + z
[From equation (2)]
=> ∠ODB = 900 + ……….... (4)
Also, ∠OEB = ∠OCE + ∠COE
[Exterior angle property]
=> ∠OEB = 900 – + z
[From equation (3)]
=> ∠OEB = 900 + …….... (5)
Also, ∠OED = ∠ODE = 900 – 
……... (6)
From equation (4), (5) and (6), we have
∠BDE = ∠BED = 900 + – (900 – )
=> ∠BDE = ∠BED = 900 + – 900 +
=> ∠BDE = ∠BED = (y + z) ÷ 2
=> ∠BDE = ∠BED = y + z ........ (7)
So, ∠BDE = 1800 – (y + z)
=> ∠ABC = 1800 – (y + z) ......... (8)
Now, (y - z) ÷ 2 = (3600 – y – 2z – y) ÷ 2 = 1800 – (y + z) …...... (9)
From equation (8) and (9), we get
∠ABC = 