Given AB and CD are two chords of circle with centre O, which intersects at E.

Construction Join AC, BC and BD.

Proof AC is a chord.

∴ ∠AOC=2∠ABC…(i)

[angle subtended at centre is double the angle subtended at circumference]

Similarly, ∠DOB = 2DCB… (ii)

[angle subtended by the chord BD]

On adding Eqs. (i) and (ii),

we get ∠AOC + ∠DOB = 2(∠ABC + ∠DCB) …(iii)

In ∆CEB,∠AEC = ∠ECB + ∠CBE

[exterior angle is equal to the sum of two opposite interior angles ]

∠AEC = ∠DCB +∠ABC ,…(iv)

[∴ ∠ECB =∠DCB and ∠CBE ∠ABC] :.

From Eqs. (iii) and (iv), we get

∠AOC + ∠DOB =2∠AEC