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Constructing a Line to Divide Area of a Triangle in Half


Date: 05/13/98 at 01:53:02
From: Nathan Rabiroff
Subject: Similar Triangles

Here is the problem I was asked to solve:

           C
           /\
          /  \
         /    \
        /      \
       /        \
      /__________\
     A            B

Can you cut triangle ABC into two pieces of equal area by drawing a 
line parallel to one of the sides? How would you find it?

I know there is a way to divide it, but I can't figure out how to do 
it. I have spent hours trying to figure it out, and desperately need 
help. I already stumped my math tutor with it, and need someone else 
to look at it.

Thanks,
Nate


Date: 05/13/98 at 11:58:36
From: Doctor Rob
Subject: Re: Similar Triangles

Drop a perpendicular from C to AB, meeting AB at D. On that 
perpendicular, mark a point P such that PC/PD = sqrt(2)/2. Construct a 
line through P parallel to AB (and perpendicular to PD), and that will 
do it. (Why? You supply the reason. It has to do with similar 
triangles having proportional sides, and with the formula for the area 
of a triangle as b*h/2.)

Here is another construction of the same line. On side AC, mark point 
A' such that A'C/AC = sqrt(2)/2. On side BC, mark point B' such that 
B'C/BC = sqrt(2)/2. Connect A' and B', and this line is the right one.

If you need to know how to construct P (or A', or B') using 
straightedge and compass, here is how. Draw the perpendicular bisector 
of the line segment CD (or AC, or BC). Using the midpoint of the 
segment as a center, draw a circle with the segment as a diameter.  
Draw a circular arc whose center is C and whose radius is the distance 
from C to the intersection of the circle and the perpendicular 
bisector. Its intersection with the segment will be P (or A', or B').

-Doctor Rob, The Math Forum
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Associated Topics:
High School Constructions
High School Geometry
High School Triangles and Other Polygons

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