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 --- 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.