QQuestion
Two chords, AB and CD of a circle with center O, when produced, intersect each other at the point P. Prove that ∠AOC−∠BOD = 2 × ∠BPC
✓Answer
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)
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