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From: leeper@mtgzz.UUCP (m.r.leeper)
Newsgroups: net.puzzle
Subject: Re: Geometry problem
Message-ID: <1417@mtgzz.UUCP>
Date: Sat, 16-Nov-85 09:32:14 EST
Article-I.D.: mtgzz.1417
Posted: Sat Nov 16 09:32:14 1985
Date-Received: Mon, 18-Nov-85 05:37:53 EST
References: <2966@brl-tgr.ARPA>
Organization: AT&T Information Systems Labs, Middletown NJ
Lines: 29

The distance from the line Px+Qy+R = 0 to the point (U,V) is

	    |PU+QV+R|/(P^2+Q^2)^(1/2) 

This can be proved separately, and the interpretation is fairly
interesting, but let us apply it here by placing a Cartesian axis over
the problem with A at (0,0) and B at (0,1).  Let the base angle at A be
2a and the base angle at B be 2b.  So the line BC has the equation

	    y + (tan(2b))x - tan(2b) = 0

The line AC has equation 

	    y - (tan(2a))x = 0

using the above formula, the length of segment AE is

	    tan(2b)/((1+(tan(2b))^2)^(1/2))

the length of segment BD is

	    tan(2a)/((1+(tan(2a))^2)^(1/2))

A little simple (careful not to lose roots) manipulation tells you that
2a = 2b.  Hence the base angles of the triangle are equal so the
triangle is isoceles.

				Mark Leeper
				...ihnp4!mtgzz!leeper