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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.

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Given : Two equal chords AD and CE of a circle with centre O. When meet at B when produced. 

To Prove : ∠ABC = 1/ 2 (∠AOC – ∠DOE) 

Proof : Let ∠AOC = x, ∠DOE = y, ∠AOD = z 

∠EOC = z [Equal chords subtends equal angles at the centre] 

∴ x + y + 2z = 36° [Angle at a point] .. (i) 

OA = OD ⇒ ∠OAD = ∠ODA 

∴ In DOAD, we have 

∠OAD + ∠ODA + z = 180° 

⇒ 2∠OAD = 180° – z [ ∠OAD = ∠OBA] 

⇒ ∠OAD = 90° – z/ 2 ... (ii) 

Similarly ∠OCE = 90° – z/ 2 ... (iii) 

⇒ ∠ODB = ∠OAD + ∠ODA [Exterior angle property] 

⇒ ∠OEB = 90° – z/ 2 + z [From (ii)] 

⇒ ∠ODB = 90° + z/ 2 ... (iv) 

Also, ∠OEB = ∠OCE + ∠COE [Exterior angle property] 

⇒ ∠OEB = 90° – z/ 2 + z [From (iii)] 

⇒ ∠OEB = 90° + z/ 2 ... (v)

Also, ∠OED = ∠ODE = 90° – y/ 2 ... (vi) 

O from (iv), (v) and (vi), we have 

∠BDE = ∠BED = 90° + z/ 2 – (90° − y/ 2)

⇒ ∠BDE = ∠BED = y/ z + 2 

⇒ ∠BDE = ∠BED = y + z ... (vii) 

∴ ∠BDE = 180° – (y + z) 

⇒ ∠ABC = 180° – (y + z) ... (viii) 

Now, (y − z)/2 = (360o - y -2z - y) / 2

= 180° – (y + z) ... (ix) 

From (viii) and (ix), we have 

∠ABC = x- y /  2 Proved

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