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# BMO1 Geometry Question watch

1. This is question 4 from BMO1 2014 and I've gotten stuck on it. I've drawn the diagram:

All it tells you is that the minor arcs AF and BF are equal and it wants you to prove that the lines AB and PQ are parallel. What I've done so far is in the attachments, I can write all of the red angles in terms of A,B,C,D and I know that A+B+C+D = 180 deg.
There's lots of similar triangles all over the place but they're really not helpful. I've thought about doing a converse corresponding angle proof with the green angles, but for that I need to know PQF. Any help would be appreciated

Let ABCD be a cyclic quadrilateral where F is the midpoint of the arc AB of it circumcircle. Let AC and FD intercept at P and BD and FC intersect at Q. Prove tha PQ and AB are parallel
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2. (Original post by BobBobson)
This is question 4 from BMO1 2014 and I've gotten stuck on it. I've drawn the diagram:

All it tells you is that the minor arcs AF and BF are equal and it wants you to prove that the lines AB and PQ are parallel. What I've done so far is in the attachments, I can write all of the red angles in terms of A,B,C,D and I know that A+B+C+D = 180 deg.
There's lots of similar triangles all over the place but they're really not helpful. I've thought about doing a converse corresponding angle proof with the green angles, but for that I need to know PQF. Any help would be appreciated

Let ABCD be a cyclic quadrilateral where F is the midpoint of the arc AB of it circumcircle. Let AC and FD intercept at P and BD and FC intersect at Q. Prove tha PQ and AB are parallel
I've had a look at it for quite a while and can't see it. If you need the solution urgently, your best option is probably to pay £15 for the UKMT 2014-15 yearbook (available at http://shop.ukmt.org.uk/ukmt-books/2014-2015-yearbook) which will include the BMO solutions. Alternatively, if you're willing to wait, you can make a Freedom of Information Act request to the University of Leeds, which the UKMT is a part of.
3. (Original post by HapaxOromenon3)
I've had a look at it for quite a while and can't see it. If you need the solution urgently, your best option is probably to pay £15 for the UKMT 2014-15 yearbook (available at http://shop.ukmt.org.uk/ukmt-books/2014-2015-yearbook) which will include the BMO solutions. Alternatively, if you're willing to wait, you can make a Freedom of Information Act request to the University of Leeds, which the UKMT is a part of.
It's alright, I don't need it urgently, I was just interested is all. It's shame they make you pay for solutions, despite them having sponsors. It seems as if I'm just missing something obvious. Normally you get stuck because you don't know what to do, but here you're stuck because you don't know what you can even do
4. I think the solution would go something like this:

Let AC and BD meet at X.
All we need to show is XBA is an enlargement of XPG.
BXC and AXD are similar
QXC and PXD can be shown to be similar quite easily, and from there you should be able to show XBA is an enlargement of XPG by considering ratios XQ:XB and XP:XA.

Might be wrong, but I think that's how I did it
5. (Original post by A02)
I think the solution would go something like this:

Let AC and BD meet at X.
All we need to show is XBA is an enlargement of XPG.
BXC and AXD are similar
QXC and PXD can be shown to be similar quite easily, and from there you should be able to show XBA is an enlargement of XPG by considering ratios XQ:XB and XP:XA.

Might be wrong, but I think that's how I did it
Yes, this works!

I wasn't sure this thread would ever get an answer. I had a go earlier but couldn't get anywhere.
6. (Original post by A02)
I think the solution would go something like this:

Let AC and BD meet at X.
All we need to show is XBA is an enlargement of XPG.
BXC and AXD are similar
QXC and PXD can be shown to be similar quite easily, and from there you should be able to show XBA is an enlargement of XPG by considering ratios XQ:XB and XP:XA.

Might be wrong, but I think that's how I did it
Thanks so much, it seems so simple now. It almost seems too simple to be a Q4.

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