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From: ags@pucc-h (Dave Seaman)
Newsgroups: net.puzzle
Subject: Geometry Problem (algebraic proof)
Message-ID: <2456@pucc-h>
Date: Sun, 17-Nov-85 13:23:17 EST
Article-I.D.: pucc-h.2456
Posted: Sun Nov 17 13:23:17 1985
Date-Received: Tue, 19-Nov-85 04:00:51 EST
Organization: Purdue University Computing Center
Lines: 52
Keywords: angle bisectors

In a triangle with sides a, b and c the length of the angle bisector which
meets side c is given by

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

where s = (a + b + c) / 2.  Note that "abs(s-c)" means "a*b*s*(s-c)" and
not the absolute value of (s-c).

I will post a proof of this formula as a separate lemma.  By symmetry,
the corresponding formula for the bisector that meets side b is

		      2
		t  = --- sqrt(acs(s-b))
		 b   a+c

Given t  = t  , we are to show that b = c.
       c    b

After clearing fractions we have

		(a+c) sqrt (abs(s-c)) = (a+b) sqrt(acs(s-b)).

Squaring both sides and using the identities s-c = (a+b-c)/2  and 
s-b = (a+c-b)/2:

		     2                2
		(a+c) abs(s-c) = (a+b) acs(s-b)
or
		  2      2              2      2
		(a +2ac+c )b(a+b-c) = (a +2ab+b )c(a+c-b)

which can be rearranged to

		 3   2 2    2   3     3   2 2     2   3
		a b+a b +3ab c+b c = a c+a c +3abc +bc 

or

		 3         2  2  2                2  2
		a (b-c) + a (b -c )+3abc(b-c)+bc(b -c ) = 0

which factors as

		       3  2
		(b-c)[a +a (b+c)+3abc+bc(b+c)] = 0.

Since a, b and c are all positive, the expression in brackets cannot be
zero.  Therefore it is possible to divide, yielding b-c=0, or b=c.
-- 
Dave Seaman			 ..!pur-ee!pucc-h!ags