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[Maths] 簡單circle

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簡單circle


(a)(i)已經睇唔到= =
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要PROVE 拒CONGRUENT?

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angle ARP=angle RFP (angle in alt. segment)
angle PRF=angle QPB (angle in alt. segment)
              =angle PAB (angle in alt. segment)
angle APR=angle BPR (angle sum of triangle)
triangle PAR~triangle PRF (AAA)

[ 本帖最後由 桃子 於 2010-3-31 08:28 PM 編輯 ]

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回覆 #3 桃子 的帖子

angle PRB=angle QPB (angle in alt. segment)    你係咪寫錯左... angles in alt. segment 要係triangle 3個vertex 掂住 circle 先用得.

[ 本帖最後由 manng825 於 2010-3-30 03:28 PM 編輯 ]

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樓主, 你有冇寫漏野,  你係邊度搵呢題出黎架?

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回覆 #4 manng825 的帖子

Sorry, that's a typo.

It should be obvious as the corresponding angle of angle PAB is angle PRF instead of angle PRB.

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a i) is the most difficult,
but if you assume a i) is correct, the others are easy.
Failing to prepare is preparing to fail!!!
想入好既大學,先要公開試考得好,更要jupas排得好!!!

想搵人補習??? Click on this:
http://www.lsforum.net/board/vie ... e%3D1&frombbs=1

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回覆 #6 桃子 的帖子

point B is not on the circle C1
Failing to prepare is preparing to fail!!!
想入好既大學,先要公開試考得好,更要jupas排得好!!!

想搵人補習??? Click on this:
http://www.lsforum.net/board/vie ... e%3D1&frombbs=1

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引用:
原帖由 peterp2509 於 2010-3-31 11:32 PM 發表
point B is not on the circle C1
For (a)(i) to hold, point B has to be on the circle C1.
Part (a)(i) is similar to a past paper question, which required this point to be on the outer circle.


In this modified version, there seems to be no restriction on angle BPR.
If we pull B to the bottom right, angle BPR is increased but the conditions in the question are still satisfied.
If we pull B into circle C1, angle BPR is decreased but the conditions in the question are still satisfied.

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