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Dividing a Line Segment

Date: 03/08/2001 at 02:14:08
From: Ekmel
Subject: Geometry - angle bisector

How do I divide a line segment into a 2:3 ratio using the angle 

So far:

  draw auxilary line
  construct parallel line
  choose a random compass setting and mark it off

Date: 03/08/2001 at 15:28:20
From: Doctor Rob
Subject: Re: Geometry - angle bisector

Thanks for writing to Ask Dr. Math, Ekmel.

Here is the diagram:

           P2 o
             / \
            /   \
        P1 o     \
          /       \
         /         \                   /
      A o-----------o-----------------o B
       /           X \               /
                      \             /
                       \           o Q1
                        \         /
                         \       /
                          \     o Q2
                           \   /
                            \ /
                             o Q3

Through A draw an auxiliary line AC. Through B draw a line BD parallel
to AC. Set the compass at some opening, and from A mark P1 on AC. From 
P1 mark P2 on AC. From B, on the opposite side of AB mark Q1 on BD.  
From Q1 mark Q2 on BD. From Q2 mark Q3 on BD. Now connect P2Q3.
The point where this intersects AB is X, which satisfies AX:XB = 2:3.

I leave it to you to prove that this works.  Hint: triangles AXP2 and
BXQ3 are similar.

- Doctor Rob, The Math Forum   

Date: 03/11/2001 at 18:17:34
From: The Lemke's
Subject: Re: Geometry - angle bisector

Thanks. I had gotten just about that far on my own - but I still do 
not see how the angle bisector fits into it. Any more hints?

Date: 03/12/2001 at 08:32:21
From: Doctor Rob
Subject: Re: Geometry - angle bisector

Thanks for writing back!

Here's another way that does use an angle bisector:

Given line segment AB, mark off from A along AB points P1, P2, and P3
such that AP1 = P1P2 = P2P3.  

With radius AP2, draw a circle with center A. With radius AP3, draw a 
circle with center B. Let one of the points of intersection of the two 
circles be C. Connect AC and BC.  

Now bisect <ACB. Let X be the intersection of that bisector and AB.  
Then AX:XB = 2:3.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry

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