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Re: ? angle bisector of triangle divides opposite side in proportion
to other 2 sides ?
Posted:
Feb 28, 1998 2:59 PM
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Charles,
John Conway's off-the-top-of-his-head trig proof suggests this geometrical
one to me.
Tri(ABZ) and TRI(CBZ) have the same altitude from B and so their area's
are in the ratio d:e.
The altitudes from Z to AB and CB are equal since they form congruent
right triangles (the triangles share BZ, the angle bisector, so they are
congruent by hy-leg)
Therefore, TRI(ABZ) and TRI(CBZ) are also in the ratio c:a.
Alan
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Alan Lipp
Williston Northampton School
Easthampton, Ma
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> 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,
> then:
> d/c = e/a.
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