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Triangle Construction

Date: 09/09/2001 at 23:24:00
From: becca
Subject: Geometry-triangles

Given a triangle ABC and point D somewhere on the triangle (not a 
midpoint or vertex), I am having trouble constructing a line that 
bisects the area.  

I have been able to construct a line parallel to one side, but it must 
be at a constant ratio (1/SQRT 2). D can be anywhere. Using Sine 
ratios and the formula for the area of the triangle, I have discovered 
that, if D lies on BC and F is the point at which the line intersects 
AB, then BA/BF = 2BD/BC. (This assumes that D intersects AB, but a 
similar ratio can be derived if it intersects AC.) But I don't see how 
this can help me to construct the line.  

I've been working on this for days. I hope you can help!  
Many thanks.

Date: 09/10/2001 at 04:05:20
From: Doctor Floor
Subject: Re: Geometry-triangles

Hi, Becca,

Thanks for writing.

Let me make a sketch to start with:

       / \
      /   \
     /     E
    /       \
    x     y

I have located D on side AB, and lengths AD = x, DB = y with x < y. 
In this case E must be on side BC. If x = y, then point E coincides 
with C, and when x > y, E should lie on AC.

Now the region ADEC can be divided into two triangles by segment DC. 
Triangle ADC has area 

 --- * K

where K is the area of ABC. This can be seen by noticing that the 
heights of triangles ADC and DBC are equal, and their bases are 
x and y respectively.

The areas of ADEC and DBE should be

  1     (x+y)/2
  - K = -------
  2       x+y

and thus triangle DEC should have area

  ------- .

So the ratio of the areas of triangles DBE and DEC should be 
x + y : y - x. Then, again because these triangles have equal heights 
seen from bases DE and EC, this should give us that 
BE : EC = x + y : y - x.

Now point A can be constructed in the following way:

                        E   AE//XC
     y-x       x+y


It is not too difficult to find point X and (since XB  is 2y, point X 
is B reflected through D). Draw XC and let E be the intersection of BC 
and the line through A parallel to XC.

If you need more help, just write back.

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

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