
? angle bisector of triangle divides opposite side in proportion to other 2 sides ?
Posted:
Feb 28, 1998 10:17 AM


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.
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

