A, B and P are the points on the circle with centre O - NSC Mathematics - Question 9 - 2017 - Paper 2
Question 9
A, B and P are the points on the circle with centre O. AO, BO, AP and BP are drawn.
Prove the theorem which states that ∠AOB = 2∠APB.
Worked Solution & Example Answer:A, B and P are the points on the circle with centre O - NSC Mathematics - Question 9 - 2017 - Paper 2
Step 1
Understanding the Configuration
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Answer
Begin by noting points A, B, and P lie on the circumference of the circle centered at O. The angles ∠AOB and ∠APB are inscribed angles where AOB is the angle at the center and APB is the angle at the circumference.
Step 2
Applying the Circle Theorems
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Answer
Recall that for any point P on the circumference of the circle, the angle subtended by a chord at the center is twice the angle subtended at any point on the circumference. Thus, by the circle theorem:
∠AOB=2∠APB
Step 3
Concluding the Proof
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Answer
Hence, this correct and formal proof affirms that the angle at the center (∠AOB) is indeed twice that of the angle at the circumference (∠APB), which is consistent with the properties of cyclic quadrilaterals.
