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9.1 In the diagram, O is the centre of a circle - NSC Mathematics - Question 9 - 2022 - Paper 2
Question 9
9.1 In the diagram, O is the centre of a circle. OD bisects chord AB. Prove the theorem that states that the line from the centre of a circle that bisects a chord ... show full transcript
Worked Solution & Example Answer:9.1 In the diagram, O is the centre of a circle - NSC Mathematics - Question 9 - 2022 - Paper 2
Step 1
Prove the theorem that states that the line from the centre of a circle that bisects a chord is perpendicular to the chord, i.e. OD ⊥ AB.
Answer
To prove that line OD is perpendicular to chord AB, we can construct points A and B in a circle centered at O. By drawing lines OA and OB, we notice that OA = OB (radii of the circle).
Next, we note that since OD bisects AB, we can set AD = DB. In the triangle OAD and OBD, we have:
- OA = OB (common sides)
- AD = DB (AD is equal to DB by construction)
- ∠ADO = ∠BDO (angles opposite equal sides)
Thus by the Isosceles Triangle Theorem, triangles OAD and OBD are congruent. Hence, ∠AOD = ∠BOD, leading to the conclusion that OD ⊥ AB.
Step 2
Prove, giving reasons, that: 9.2.1 OTBG is a cyclic quadrilateral
Answer
To prove that OTBG is a cyclic quadrilateral, observe that we have:
- ∠OTB = 90° (the angle from the center to the midpoint of chord)
- ∠OGB = 90° (tangent line at B by the properties of tangents)
- Therefore, ∠OTB + ∠OGB = 90° + 90° = 180°.
Since the opposite angles sum to 180°, OTBG is confirmed to be a cyclic quadrilateral.
Step 3
Prove, giving reasons, that: 9.2.2 ∠GOB = ∠Ŝ
Answer
By the property of angles in the same segment, since G lies on the circumference of the circle, thus line segments GB and OS create equal angles with respect to points in the same segment. Therefore, we have:
∠GOB = ∠Ŝ.
This demonstrates that the angles are equal based on the inscribed angle theorem.