Hi,

Encountered this question, but don't know how to go about proving it. If anyone could help, that would be much appreciated!

Oh boy, love a good circle proof. Everything I'm spitting out is literally me working through this question in real time - that means at some point I may hit a block or have to disregard something entirely. I think it's important to SHOW this process, because a lot of students will often ask, "but how did you know to try x?" - the true, honest to goodness, answer is - I didn't know to try x, I just did it, and it happened to work out.

Okay, so the question is - how can we prove this? Well, we know that all tangents to a circle meet the radius at 90 degrees - so maybe we can prove that each PQ and PR are parrallel to the respective tangents at the points PQ and PR cross on the circle edge. BUT, that would only work if P is the middle of the circle - which isn't necessarily true.

Okay, instead of questioning how to answer the question, let's instead just figure out what we know and work from there. So, <ADC and <ABC must add to 180 degrees, and the same for <DAB and <DCB. But, I'm not sure how we could use this... But it does mean we know the angles <QBC and <RDC. It's worth noting down, even if we never use it. Let's say <ADC=<QBC=d, and <ABC=<RDC=b.

In fact, we can use this to figure out all the angles in the triangle. Using the same logic, <RCD=<QCD=<DAB=a (again, I've just decided that it should be called a, because that makes things easier). So that means that we now know all the angles in the two triangles they made up, RDC and QBC. In fact, now we know:

<DRC=180-b-a

<BQC=180-d-a

So, how does that help us? Well, let's call the intersection between QD and RP, M, and the intersection between RB and QP, N (I hope you're labelling this on your own circle), then all we need to prove is that either <MQP + <QMP = 90 degrees OR <NRP + <RNP = 90 degrees. And, using the above, we know that:

<DRC = 2*<NRP = 180 - b - a, <NRP = 90 - (b+a)/2

<BQC = 2*<MQP = 180 - d - a, <MQP = 90 - (d+a)/2

So for simplicity, I'm just going to work on <PNR. So I know for <NRP + <PNR = 90, then I need <PNR = (b+a)/2 - the question is, how can I get there? Well, I know that <PNR=<BNQ=180-<CNQ=180-<PNB. Can I figure any of those out?

The problem with any of these triangles is that N doesn't lie on the circle - it's inside the circle, and we don't have many proofs that rely on working INSIDE the circle - so frustratingly, knowing we're so close, this approach looks doomed to fail.

Okay, new idea. What if we made a circle using the points P, Q, and R. Then, all we'd need to do is prove that the length QR passes through the centre of that circle. Then the angle <QPR HAS to be 90 degrees.

... Except I'm not sure how we could use any of the information we've been given to inform ourselves of this circle we just made. Okay, maybe I'll go back to the (a+b)/2 thing

Okay, so let's try a slightly different approach, and we're going to extend QP so that it intersects with AR - and we'll call that point Z because I'm running out of letters that make sense. Since RP bisects the triangle ZRN, if PQ and PR are perpendicular, then <RZN=<RNZ. No, I'm stopping this train of thought, things are just going to get worse before they get better. There's gotta be a simpler way.

Okay, I think I've got it. If I go back to the (a+b)/2 thing, I know that <QNB=<RNP, and I want to show that <RNP=(a+b)/2. I may have realised this earlier if I labelled myself (oops), but I just realised that I CAN figure out <QNB because I know all the other angles in the triangle QNB:

<BQN=<MQP=90-(d+a)/2

<NBQ=d

<QNB=180-<BQN-<NBQ=180-(90-d/2-a/2)-d=90+a/2-d/2=90+(a-d)/2

Okay, not looking good. But, we know that d=180-b, so we then get:

<RNP=<QNB=90+a/2 - (180-b)/2=90 + a/2 - 180/2 + b/2 = 90 + a/2 - 90 + b/2 = (a+b)/2

Which means that:

<RPQ=<RPN=180-<RNP-<NRP=180-(a+b)/2-(90-(a+b)/2)=90

Fuck, this was a work-out. I'm sure this got confusing along the way, so try re-doing this yourself to see if it makes more sense

EDIT: After giving up I hit a break-through, so hopefully you've noticed this and come back after I made my initial post? Oops