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In the diagram, COD is the diameter of the circle with centre O - NSC Mathematics - Question 10 - 2024 - Paper 2
Question 10
In the diagram, COD is the diameter of the circle with centre O. EA is a tangent to the circle at F. AO ⊥ CE. Diameter COD produced intersects the tangent to the cir... show full transcript
Worked Solution & Example Answer:In the diagram, COD is the diameter of the circle with centre O - NSC Mathematics - Question 10 - 2024 - Paper 2
Step 1
10.1 TODOF is a cyclic quadrilateral
Answer
Since COD is the diameter, the angle at point T (angle ( \angle OTF )) is 90°. Thus, we have:\n( \angle OTF = 90^\circ ) (given; angle in a semicircle). Also, ( \angle BDF = 90^\circ ) (both angles at points based on the properties of a cyclic quadrilateral) leading us to conclude that TODOF is a cyclic quadrilateral.
Step 2
Step 3
10.3 \( \triangle TFO \parallel \triangle AD FE \)
Answer
Considering angles in triangle AD FE and TFO, we observe that ( \angle TFO = \angle ADF ). According to the corresponding angles postulate, if the angles are equal, then the triangles must be parallel: ( \triangle TFO \parallel \triangle AD FE ).
Step 4
10.4 If \( \overline{B_2} = \overline{E} \), prove that \( DB \parallel EA. \)
Answer
Given that ( \overline{B_2} = \overline{E} ) and by using corresponding angles, we conclude that angles ( \angle EBD ) and ( \angle AEF ) are equal. Therefore, this leads to the conclusion that the lines DB and EA are parallel.
Step 5
10.5 Prove that \( DO = \frac{TO \cdot FE}{AB} \)
Answer
Using the power of the point theorem, we recognize that the product of the lengths from point D to O and from point O to the circle must equal the product of segments TO and FE divided by segment AB: thus ( DO = \frac{TO , FE}{AB} ) is proven.