(i)

Angle PRQ, because it is an angle subtended by the chord PQ where P is a tangential point of the chord ST to the circle

[In short, we are talking about the tangent-chord theorem or the alternate segment theorem]

(ii)

Because angles PMR and PNR are both equal to 90 degrees and they would be angles in a semicircle, so conversely, M and N must lie along the circumference of this circle

(iii)

To do this, we simply prove that angles RMN and RQP are the same (and then talk about corresponding angles) or angles MNT and QPT are the same (and then talk about corresponding angles) or something similar.

Ok, this question is particularly difficult for most of us because of the complexity of the question [I was lucky I noticed something early on or I would also be equally stuck]

Parts i and ii, as it turns out, are largely crucial to part iii - not because simply "we can draw a circle to pass through points M and N', but because for some reason I felt that they would not have asked parts i and ii randomly for no reason.

Here's the deal.

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If a circle can be drawn to pass P, R, M and N with PR being the diameter, then

• consider PM as the chord
• PNM = PRM because they are angles lying within the same segment

But PRM is obviously PRQ and I already noted in part i that PRQ = QPT, so...

PRM = QPT
PNM = QPT

and it's now obvious that

TNM = TPQ

Focusing on these, we can conclude, conversely, that lines MN and QP are parallel because TNM and TPQ satisfy the property of corresponding angles.

Thank you