Two chords AB and CD of a circle, centered at O, intersect each other internally at point P. Prove that \(\angle\)AOD + \(\angle\)BOC = 2\(\angle\)BPC.

Q.Two chords AB and CD of a circle, centered at O, intersect each other internally at point P. Prove that \(\angle\)AOD + \(\angle\)BOC = 2\(\angle\)BPC.

In the given circle with center O, chords AB and CD intersect at point E. We need to prove that: ∠AOD + ∠BOC = 2∠BPC **Construction:** Join points A and D. **Proof:** The central angle ∠AOD and the inscribed angle ∠ACD subtended by the arc AD satisfy: ∴ ∠AOD = 2∠ACD --- (i) Similarly, the central angle ∠BOC and the inscribed angle ∠BAC subtended by the arc BC satisfy: ∴ ∠BOC = 2∠BAC --- (ii) Now, in triangle APC: ∠BPC = ∠ACP + ∠PAC [∵ An exterior angle is equal to the sum of the two opposite interior angles] Adding equations (i) and (ii), we get: ∴ ∠AOD + ∠BOC = 2∠ACD + 2∠BAC = 2(∠ACD + ∠BAC) = 2(∠ACP + ∠PAC) = 2∠BPC (Proved)