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#1 2007-02-19 09:24:47
- efmchico
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- Registered: 2007-01-15
- Posts: 7
geometry problem using ceva's theorem
1) Prove: if p and q are points on sides AB and AC, respectively, of trianlge ABC so that PQ is parallel to BC and if X is the point of intersection of BQ and CP then AX goes through the midpoint of BC.
The hint is to use Ceva's Theorem. I'm not sure how to start. Any help is appreciated.
Last edited by efmchico (2007-02-19 17:32:48)
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#2 2007-02-19 10:28:24
- Zhylliolom
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- Registered: 2005-09-05
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Re: geometry problem using ceva's theorem
Note that ABC and APQ are similar triangles, so that BP/PA = QC/AQ, and (BP/PA)(AQ/QC) = 1. Note that since the cevians are concurrent in this situation, Ceva's theorem may be used. Then 1 = (BP/PA)(AQ/QC)(CM/MB) = 1*(CM/MB) = CM/MB, which implies CM = MB, which tells us M is the midpoint of BC. I hope this is clear enough for you.
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#3 2007-02-19 13:45:13
- efmchico
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- Registered: 2007-01-15
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Re: geometry problem using ceva's theorem
how did you go from the step: BP/PA = QC/AQ (which is just similar triangle ratios) to
(BP/PA)(AQ/QC) = 1?
Thanks
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#4 2007-02-19 14:05:36
- Zhylliolom
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- Registered: 2005-09-05
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Re: geometry problem using ceva's theorem
Multiply each side of the equation BP/PA = QC/AQ by AQ/QC.
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#5 2007-02-19 14:12:26
- efmchico
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- Registered: 2007-01-15
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Re: geometry problem using ceva's theorem
alright, thanks. I think I understand this now.
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#6 2007-02-22 15:08:37
- NanWei
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Re: geometry problem using ceva's theorem
Hi, I do not understand. Who do you make(BP/PA)(AQ/QC)(CM/MB)=1? I know (BP/PA)(AQ/QC)=1, but what about (CM/MB)?
#7 2007-02-23 01:32:13
- mathsyperson
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- Registered: 2005-06-22
- Posts: 4,900
Re: geometry problem using ceva's theorem
(BP/PA)(AQ/QC) ≠ 1.
You can cancel the Ps and Qs in each respective term, which gets you (B/A)(A/C), then cancel the As to get B/C.
The final term can be reduced to C/B, and then multiplying this by B/C gets you 1.
Why did the vector cross the road?
It wanted to be normal.
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#8 2007-02-23 14:44:02
- John E. Franklin
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- Registered: 2005-08-29
- Posts: 3,588
Re: geometry problem using ceva's theorem
Ceva came up with this in the year my favorite musician was born, 1678 (A. Vivaldi).
But Ceva was using Menelaus's Theorem from centuries before that.
I don't really understand why it works, but here is a link:
http://agutie.homestead.com/files/ceva.htm
igloo myrtilles fourmis
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