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CTK Exchange
Ron Olszewski
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Apr-26-04, 10:30 AM (EST) |
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"Pascal's Theorem"
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I've been examining this page:
https://www.cut-the-knot.org/Curriculum/Geometry/Pascal.shtmlI understand most of the proof of Pascal's Theorem but this part has me stumped:
It'says the multiply the three identities and rearrange the terms, and the three terms in parenthses then equal 1. Why? I've studied it for a while and can't make any sense out of why the terms (ID*IE/IB*IC), (HF*HA/HD*HE) and (GC*GB/GA*GF) equal 1. I apologize if this is not the place to post this, or if the answer is extremely obvious  . -Ronald Olszewski |
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alexb
Charter Member
1264 posts |
Apr-26-04, 10:42 AM (EST) |
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2. "RE: Pascal's Theorem"
In response to message #0
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>It'says the multiply the three identities and rearrange the
>terms, and the three terms in parenthses then equal 1. Why?
This is because of the Intersecting Chords theorem. E.g., for the point I, since the chords BC and DE intersect in I, the theorem gives:ID·IE = IB·IC. >I apologize if this is not the place to post this, or if the
>answer is extremely obvious . Thank you for bringing this to my attention. I have inserted a link to the theorem right at the Pascal page. |
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