In the diagram, O is the centre of the circle - NSC Mathematics - Question 8 - 2023 - Paper 2

We can find ∠L̅ using the 'tangent-chord theorem', which states that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Since K is a point on the circle and ST is a tangent, we establish:

∠L̅ = S∠K = 36°.

To find ∠K̅O̅T̅, note that the angle at the center is double that at the circumference:

∠K̅O̅T̅ = 2 imes ∠L̅ = 2 imes 36° = 72°.

To prove KM = ML, we use the property of radius (OK) being perpendicular to the chord (KL). Thus:

In triangle KMO, the sum of internal angles gives:

to find KM: ∠KMO = 180° - (S∠K + ∠OKT) = 180° - (36° + 72°) = 72°.

This establishes that KM = ML, as they are both radius lines creating equal angles at the center.

To show that BC is parallel to AD, we can use the converse of the Basic Proportionality Theorem (also known as Thales' theorem). Since AB || DS, we have the segments divided proportionally. Given that DC = 20 and CS = 12, we check:

DC : CS = 20 : 12 = 5 : 3.

Since AB || DS, we conclude BC || AD.

Using the ratio we established earlier, we can express AD in terms of RD.

Using the information provided:

AB : BS = 5 : 3, and

RD = AR + AD, Using proportional sides, AD = 18, and AB = 20, giving AD : AB = 18 : 20 = 9 : 10.