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In the diagram, ABCD is a cyclic quadrilateral such that AC ⊥ CB and DC = CB - English General - NSC Mathematics - Question 10 - 2018 - Paper 2
Question 10
In the diagram, ABCD is a cyclic quadrilateral such that AC ⊥ CB and DC = CB. AD is produced to M such that AM ⊥ MC. Let B̂ = x . 10.1 Prove that: 10.1.1 MC is a t... show full transcript
Worked Solution & Example Answer:In the diagram, ABCD is a cyclic quadrilateral such that AC ⊥ CB and DC = CB - English General - NSC Mathematics - Question 10 - 2018 - Paper 2
Step 1
Prove that: MC is a tangent to the circle at C
Answer
To prove that MC is a tangent to the circle at C, we utilize the property that the angle subtended by a tangent at the point of contact is equal to the angle subtended by the chord through that point on the opposite arc.
Given angle ÂC = 90° − x, we know the corresponding angle in the triangle ACB is equal to angle ÂC = 90° − x, as both angles subtend the same arc AC. Hence, by the converse of the tangent theorem, the line segment MC must indeed be tangent to the circle at point C.
Step 2
Prove that: △ACB || △ACMD
Answer
In triangles ACB and ACM, since MC is a tangent and we have already established that ÂC = ÂM = 90° − x, we can apply the concept of corresponding angles.
Since angle C in triangle ACB corresponds to angle M in triangle ACMD, we conclude that:
△ACB || △ACMD.
Step 3
Hence, or otherwise, prove that: CM² = AM
Answer
By applying the properties of similar triangles derived from the previous part, particularly the ratio of the corresponding sides, we find that:
CM² = AM × MD.
Since MD equals DC by the properties of the cyclic quadrilateral, we have:
CM² = AM × DC = AB × MD,
Thus proving CM² = AM.
Step 4
Prove that: AM / AB = sin² x
Answer
Using the definition of sine in right triangle AMC, we can express the ratio AM/AB in terms of x.
By the definition of sine, we have:
AM = AC × sin x AB = AC,
Thus,
AM / AB = AM / AC = sin² x.
Therefore, we arrive at the desired conclusion that AM / AB = sin² x.