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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




Re: ? angle bisector of triangle divides opposite side in proportion to other 2 sides ?
Posted:
Feb 28, 1998 2:59 PM


Charles,
John Conway's offthetopofhishead 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 hyleg)
Therefore, TRI(ABZ) and TRI(CBZ) are also in the ratio c:a.
Alan  Alan Lipp Williston Northampton School Easthampton, Ma 
> 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.



