In the diagram, O is the centre of circle ABCD and BOD is a diameter - NSC Mathematics - Question 9 - 2017 - Paper 2
Question 9
In the diagram, O is the centre of circle ABCD and BOD is a diameter. F, the midpoint of chord AC, lies on BOD. G is a point on AD such that GO ⊥ DB.
9.1 Give a rea... show full transcript
Worked Solution & Example Answer:In the diagram, O is the centre of circle ABCD and BOD is a diameter - NSC Mathematics - Question 9 - 2017 - Paper 2
Step 1
9.1.1 Give a reason why: DÂB = 90°
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Answer
The angle DÂB is a right angle because it is inscribed in a semicircle. According to the inscribed angle theorem, an angle subtended by a diameter of a circle is always a right angle.
Step 2
9.1.2 Give a reason why: AGOB is a cyclic quadrilateral
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Answer
AGOB is a cyclic quadrilateral because the opposite angles Ĝ1 and B̂1 are supplementary. This is due to both angles lying in the same segment of the circle.
Step 3
9.2.1 Prove that: AC || GO
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Answer
To prove that AC is parallel to GO, we note that both lines subtend equal angles at the points O and G. Therefore, by the Alternate Interior Angles Theorem, we conclude that AC || GO.
Step 4
9.2.2 Prove that: Ĝ1 = B̂1
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Answer
Given that Ĝ1 and B̂1 subtend the same arc AB in the circle, it follows that Ĝ1 = B̂1 by the property of angles in the same segment of a circle.
Step 5
9.3 If it is given that FB = \frac{2}{5} r, determine, with reasons, the ratio of \frac{DG}{DA}
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Answer
Since FB = \frac{2}{5} r, we note that the segments DF and DG can be expressed relative to FB. Using the proportionality of segments, we find that \frac{DG}{DA} = \frac{r - FB}{r} = \frac{r - \frac{2}{5} r}{r} = \frac{3}{5}. Therefore, the ratio of \frac{DG}{DA} is \frac{3}{5}.
