#### Question

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.

#### Solution

In ΔAOD and ΔCOE,

OA = OC (Radii of the same circle)

OD = OE (Radii of the same circle)

AD = CE (Given)

∴ ΔAOD ≅ ΔCOE (SSS congruence rule)

∠OAD = ∠OCE (By CPCT) ... (1)

∠ODA = ∠OEC (By CPCT) ... (2)

Also,

∠OAD = ∠ODA (As OA = OD) ... (3)

From equations (1), (2), and (3), we obtain

∠OAD = ∠OCE = ∠ODA = ∠OEC

Let ∠OAD = ∠OCE = ∠ODA = ∠OEC = *x*

In Δ OAC,

OA = OC

∴ ∠OCA = ∠OAC (Let *a*)

In Δ ODE,

OD = OE

∠OED = ∠ODE (Let *y*)

ADEC is a cyclic quadrilateral.

∴ ∠CAD + ∠DEC = 180° (Opposite angles are supplementary)

*x* + *a* + *x* + *y* = 180°

2*x* + *a* + *y* = 180°

*y* = 180º − 2*x* − *a* ... (4)

However, ∠DOE = 180º − 2*y*

And, ∠AOC = 180º − 2*a*

∠DOE − ∠AOC = 2*a* − 2*y *= 2*a *− 2 (180º − 2*x* − *a*)

= 4*a* + 4*x* − 360° ... (5)

∠BAC + ∠CAD = 180º (Linear pair)

⇒ ∠BAC = 180º − ∠CAD = 180º − (*a* + *x*)

Similarly, ∠ACB = 180º − (*a* + *x*)

In ΔABC,

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

∠ABC = 180º − ∠BAC − ∠ACB

= 180º − (180º − *a* − *x*) − (180º − *a* −*x*)

= 2*a* + 2*x *− 180º

= 1/2[4*a* + 4*x* − 360°]

∠ABC = 1/2[∠DOE − ∠ AOC] [Using equation (5)]