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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 Conway Posts: 2,238 Registered: 12/3/04
Re: ? angle bisector of triangle divides opposite side in proportion to other 2 sides ?
Posted: Feb 28, 1998 11:52 AM

On 28 Feb 1998, 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----->|
>
> Given an _arbitrary_ triangle ABC with AZ bisecting angle ABC, and lengths
> labeled as shown,
> length of AB = c,
> length of BC = a,
> length of AC = b,
> length of AZ = d,
> length of CZ = e;
> then:
> d/c = e/a.
>

Here's a quick proof using a tiny bit of trigonometry. One formula
for the area of a triangle is " 1/2 a b sin(C) ". Apply this to the
two smaller triangles of your figure with "C" being their topmost
angles (which are 1/2 of your "B"), and you'll see that their areas
are in the proportion of your c to your a. But using the better
known formula (area = 1/2 base times height) you can also see that
their areas are in the proportion of your d to your e.

That's just "off the top of my head", as they say. I have to
dash now, but will try to think of a purely geometrical proof today.

John Conway

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