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 center.
 

Let ∠ ABC = x, ∠ AOC = y and ∠ DOE = z.              …. (1)
To prove: x = 
∠ C'OD + ∠ A'OE = z - y
Let ∠ C'OD = θ
Then ∠ A'OE = z - y - θ               ...(1)     (∵ From(1))
∠ AOD = π -(∠ AOC + ∠ C'OD) = π -(y + θ )
∠ COE = π -(∠ C'OA' + ∠ A'OE)
= π - (y + z – y - θ )
= π -(z - θ )
∴ AD = CE
∴ ∠ AOD = ∠ COE                (∵ Equal chords subtend equal angles at the centre)
∴ π -(y + θ ) = π -(z - θ )
⇒ y + θ = z - θ
⇒ 2θ = z - y
⇒ θ = ∴ ∠ C'OD =
and ∠ A'OE = z - y - =
∴ ∠ AOD = π - (y + θ )
= π -
= ∠ COE
In Δ OAD,
∴ OA = OD             (∵ Radii of the same circle)
∴ ∠ OAD = ∠ ODA    (∵ Angles opposite to the same sides of a triangle are equal)
In ΔOAD,
∠ OAD + ∠ ODA + ∠ AOD = π       (∵ Sum of all the angles of a triangle is π radians)              
           ⇒ ∠ OAD + ∠ OAD + π - = π
           ⇒ 2∠ 0AD =
           ⇒ ∠ 0AD =
Similarly, ∠ OCE =
∴ ∠ OAB = π -
and ∠ OCB = π -
In quadrilateral AOCB,
                 ∠ ABC + ∠ OAB + ∠ OCB + ∠ AOC = 2π         (∵ Sum of all the angles of a quadrilateral is 2π radians)
             ⇒ x + π - + π - + y = 2π
             ⇒ x + y =
             ⇒ 2x + 2y = y + z
             ⇒ 2x = z – y
             ⇒ x =
             Hence the result.