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Topic: ? angle bisector of triangle divides opposite side in proportion to other 2 sides ?
Replies: 4   Last Post: Mar 4, 1998 10:46 AM

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 John & Teresa Egge Posts: 2 Registered: 12/10/04
Re: ? angle bisector of triangle divides opposite side in proportion to other 2 sides ?
Posted: Mar 4, 1998 10:46 AM
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Eileen Stevenson wrote:
>
> I can't seem to prove this:
>
> B
> .
> /|\
> / | \
> c / | \ a (This is not supposed to be isosceles
> / | \ This will make a better picture viewed via a
> / | \ fixed width font such as Courier New. )
> A/_____|_____\C
> d Z e
> |<-----b----->|
>

I'm afraid it's been awhile since my high school geometry, too, but it
seems to me the key to proving it is in the fact that the angle ABZ is
equal to the angle ZBC, and that both triangles share a common (equal)
side BZ. Hope this helps!
Teresa E.

> then:
> d/c = e/a.
>
> Please help. I am tutoring a bright HS student who has this as a
> postulate in his text. It has been too many years since all those college
> math courses I took...
>
> Thanks in advance,
> Charles Stevenson
> eks@worldpath.net
> stevenc@hlthsrc.com

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Date Subject Author
2/28/98 Eileen Stevenson
2/28/98 John Conway
2/28/98 Alan Lipp
3/2/98 Ken.Pledger@vuw.ac.nz
3/4/98 John & Teresa Egge

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