Hard Light Productions Forums
Off-Topic Discussion => General Discussion => Topic started by: kasperl on January 19, 2005, 08:33:18 am
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You have 2 circles, with an equal radius. You name the intersections of the circles A and B. Line l goes through A and both circles. Intersecting points with the circels are P and Q. Prove that PQB has two equal sides.
Image:
(http://www.huiswerksite.nl/mathproof.GIF)
This has been driving me nuts for a week or so now, and I haven't found anyone who could figure this out. I'm not asking my current math teacher because I'd like to avoid him if there is any way of doing so, and I'm running out of other people to ask. Any ideas?
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Calculate the angles and find if it's equilateral/right-angled then use that?
Sorry, not done geometry for 4 years..........
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And you don't get any numbers to help you, just what you wrote down? I just want to be sure :p
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Ok, I found something, but I don't know the English math terms so I'll try anyway:
The AQB angle "sees" the AB arc (the big one), AQB is the complementary of any angle whch "sees" the AB arc (the small one)
The QPB angle "sees" the AB arc (the small one), so QPB is complementary to angle AQB, and angle BQP is complementary to angle AQB, so:
angle QPB = angle BQP, so PB=BQ
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T: Off course no numbers, that's the fun of it :p (This is Wiskunde B2, 5 VWO, in case you care.)
Skippy: Thanks, I don't entirely get it, but I'll see what I can conjure up using what you said.