
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

