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From: luong@tonto.DEC (Van Luong Nguyen  UHO  DTN 264-6560)
Newsgroups: net.math
Subject: Solution to Geometry V
Message-ID: <2636@decwrl.UUCP>
Date: Tue, 17-Jul-84 12:02:48 EDT
Article-I.D.: decwrl.2636
Posted: Tue Jul 17 12:02:48 1984
Date-Received: Wed, 18-Jul-84 03:24:56 EDT
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>Subject: Geometry V
>Once again, I have another problem, although could be one of the last.  Here
>goes:  In the obtuse triangle ABC (angle C is obtuse), point M is on AB such
>that AM = MB, point D is on BC such that MD is prpndclr to BC, and point E
>is a point on ME such that EC is prpndclr to BC.  Given the area of triangle
>ABC is 24, then find area of triangle BED.  Good luck!
>Billy Pizer

There must be a mistyping in the above problem:

As stated, point E is indeterminate... The following solution ASSUMES that
E is a point on MA such that EC is prpndclr to BC.

Then, denoting by (XYZ) the area of a triangle XYZ:
	(MCB) = 0.5 BC.MD     and     (ECB) = 0.5 BC.EC
therefore:
	(MCB)/(ECB) = MD/EC
		    = DB/CB    (since MD and EC are parallel)         (1)
Draw perpendiculars CC' and DD' to EB.
Now:
	(DEB) = 0.5 EB.DD'    and     (ECB) = 0.5 EB.CC'
Therefore:
	(DEB)/(ECB) = DD'/CC'
		    = DB/CB    (since DD' and CC' are parallel)	       (2)

Combining (1) and (2) shows that:
		(DEB) = (MCB)						(3)

But (MCB) = 0.5 CC'.MB
    (MCB) = 0.5 CC'.(AB/2) = (0.5 CC'.AB)/2 = (ABC)/2 = 12              (4)

Finally, combining (3) and (4):

		(DEB) = 12  
		==========
	        

Van Luong Nguyen,
Digital Equipment Corporation.