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From: ags@pucc-h (Dave Seaman)
Newsgroups: net.puzzle
Subject: Lemma:  Length of bisector as function of sides
Message-ID: <2457@pucc-h>
Date: Sun, 17-Nov-85 13:50:43 EST
Article-I.D.: pucc-h.2457
Posted: Sun Nov 17 13:50:43 1985
Date-Received: Tue, 19-Nov-85 04:01:03 EST
Organization: Purdue University Computing Center
Lines: 73
Keywords: angle bisector, geometry problem

I posted a proof for the geometry problem which depended on a formula
for the length of a bisector as a function of the sides.  This is the
derivation of that formula.

Let a triangle have sides a, b and c.  Let t be the length of the bisector
which meets side c.  Let s = (a + b + c) / 2.  Let u be either of the two
small angles formed by the angle bisector.

The triangle with sides a, b and c has an angle 2u between sides a and b.
Its area is therefore 

		A = (1/2) ab sin 2u.

The two small triangles formed by the bisector have areas

		A1 = (1/2) at sin u
and
		A2 = (1/2) bt sin u

where t is the length of the bisector.  Since A = A1 + A2, we have

	(1/2) ab sin 2u = (1/2) at sin u + (1/2) bt sin u
or
	ab sin u cos u  = (1/2) (a+b) t sin u

which yields

			  2ab cos u
		      t = ---------.					(1)
			    a + b

Applying the law of cosines to the large triangle:

					2  2  2
		     2		       a +b -c
		2 cos u - 1 = cos 2u = --------
					  2ab


					2      2  2
				  2    a +2ab+b -c
			     2 cos u = ------------
					    2ab

					    2  2
				  2    (a+b) -c
			       cos u = ---------
					  4ab

				       a+b+c   a+b-c    1
				     = ----- * ----- * --
					 2       2     ab

				       s(s-c)
				     = ------
					 ab
Substituting in (1):

		     2
		t = --- ab cos u
		    a+b

		     2          / s(s-c) \
		  = --- ab sqrt | ------ |
		    a+b         \   ab   /

		     2
		  = --- sqrt(abs(s-c))
		    a+b

This was the formula used in the proof.
-- 
Dave Seaman			 ..!pur-ee!pucc-h!ags