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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
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Constructions
High School Geometry
High School Triangles and Other Polygons

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