**Data :** The vertex of angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle.

**To Prove:** ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre. OR

∠ABC= \(\frac{1}{2}\)[∠DOE – ∠AOC].

**Construction:** OA, OC, OE, OD are joined.

**Proof:** In ∆AOD and ∆COE,

OA = OC,

OD = OE radii of same circle.

AD = CE (Data)

∴ ∆AOD ≅ ∆COE (SSS Postulate)

∠OAD = ∠OCE ……….. (i)

∴∠ODA = ∠OEC …………. (ii)

OA = OD

∴ ∠OAD = ∠ODA …………. (iii)

From (i) and (ii),

∠OAD = ∠OCE = ∠ODA

= ∠OEC = x°.

In ∆ODE,

OD = OE

∠ODE = ∠OED = y°.

ADEC is a cyclic quadrilateral.

∴ ∠CAD + ∠DEC = 180°

x + a + x + y = 180

2x + a + y = 180

y = 180 – 2x – a ……….. (iv)

But, ∠DOE = 180 – 2y