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.

In ΔAOD and ΔCOE,

OA = OC (radii)

OD = OE (Radii)

AD = CE (Given)

Therefore,

By SSS congruence,

ΔAOD ΔCOE

∠OAD = ∠ OCE (By CPCT) (i)

∠ODA = ∠OEC (By CPCT) (ii)

∠OAD = ∠ODA (iii)

Using (i), (ii) and (iii), we get

∠OAD = ∠OCE = ∠ODA = ∠OEC = x

Now,

In triangle ODE,

OD = OE

Hence, ABCD is a quadrilateral

∠CAD + ∠DEC = 180^o

a + x + x + y = 180^o

2x + a + y = 180^o (iv)

∠DOE = 180^o – 2y

And,

∠COA = 180^o - 2a

∠DOE - ∠COA = 2a – 2y

= 2a – 2 (180^o - 2x – a)

= 4a + 4x – 360^o (v)

∠BAC = 180^o – (a + x)

And

∠ACB = 180 – (a - x)

In ΔABC,

∠ABC + ∠BAC + ∠ACB = 180^o (Angle sum property of triangle)

∠ABC = 2a + 2x – 180^o

∠ABC = 1/2 (4a + ax – 360⁰) [using (v)]