Let AB and CD be two chords of a circle with centre O. When produced, AB and CD intersect at P.

To prove: ∠AOC − ∠BOD = 2 ∠BPC

Construction: Join B and C.

Proof: The central angle produced by arc AC is ∠AOC, and the angle in the circle subtended by the same arc is ∠ABC.

Therefore, ∠AOC = 2 ∠ABC …….. (1) [by theorem]

Similarly, the central angle produced by arc BD is ∠BOD, and the angle in the circle subtended by the same arc is ∠BCD.

Therefore, ∠BOD = 2 ∠BCD …….. (2)

Now, subtracting (2) from (1), we get:

∠AOC − ∠BOD = 2 ∠ABC − 2 ∠BCD …….. (3)

Again, in triangle BPC, external angle ∠ABC = ∠BPC + ∠BCP.

But here, ∠BCP = ∠BCD.

So, ∠ABC = ∠BPC + ∠BCD.

Multiplying both sides by 2,

2 ∠ABC = 2 ∠BPC + 2 ∠BCD …….. (4)

From (3) and (4),

∠AOC − ∠BOD = 2 ∠BPC (Hence proved)