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**Half the Area of a Triangle: A Line Parallel to the
Base**

*Problem
1: For any triangle, construct a
segment parallel to a base of the triangle that divides the triangle into two
equal areas.*

Before we start constructing,
letŐs see if we can find a ratio of the corresponding sides of the similar
triangles where the areas are in ratio of 1:2.

Let the triangle
be labeled as follows:

Because PQ is
parallel to BC, triangle AQP is similar to triangle ABC. The coefficient of similitude gives the
equality

The relationship
of the areas of the two triangles gives

Dividing both
sides by h_{2} gives

By substitution,

Now we can see the ratio of
the sides. LetŐs construct a
segment of length m parallel to the base of length k to divide the area of the
triangle in half.

Now we are ready
to begin constructing. Take any
triangle ABC. Construct a segment
B'C of length BC and perpendicular to BC at C. Draw B'B.

Find the midpoint
M of BB'. Mark an arc of radius MB with center at B to intersect the segment BC
at point N.

Now, construct a
line parallel to AB through N. This line will intersect AC at P and a parallel
line to BC through P will intersect side AB at Q. PQ is the desired segment
that divides the original triangle into equal areas.

The area of AQP
is half the area of ABC. Notice also that the area of
quadrilateral BQPC must equal to the area of AQP.

Follow this link for a GSP
construction, which also contains a GSP tool that you can use.

*Problem 2,
Part A: If the parallel segment that divides the triangle into two equal areas
is drawn for each base, a smaller triangle is formed. What is the ratio of the
area of the small triangle to the original?*

Since we have
parallel lines, then by AAA, the small yellow triangle is similar to triangle
ABC. By investigations in GSP, I
have found that the ratio between the small triangle and the large triangle is
approximately 0.01472. Click this
GSP link to investigate for yourself.

*Problem 2,
Part B: What is the ratio of the area of the shaded triangle to the area of the
original triangle in the figure below? Here again the segments parallel to the
bases divide the original triangles into two equal areas.*

Again, by AAA,
the yellow shaded triangle is similar to the original triangle. By investigations in GSP, I have found
that the ratio between the shaded triangle and the large triangle is
approximately 0.1716. Click this
GSP link to investigate for yourself.

*Problem 2,
Part C: Prove that the measures of the three shaded areas in each of the
figures below are the same. In each figure what is the ratio of the area of one
of the regions to the area of the original triangle?*

LetŐs label each of the
figure spaces A-G as follows:

Thus, by the principle of
halving the area, we know that:

By regrouping and simplifying, we find that (1):

This can be represented in
the following picture:

To prove that A=E=G, we must
first see that they are all parallelograms. Since the ½ areas are made by lines parallel to the
base, then they are obviously parallelograms. LetŐs label the points of our picture.

We know that XR = SY since RS
|| XY, TY = UZ since TU || YZ, and WZ = XV since VW || XZ. Therefore, figures A, E, and G are
actually rhombuses with equal area (A = E = G). By substitution into our previous equations (1), we see that
B = D = F as well. Therefore, we
have proven the three shaded areas in the figures are equal.

The ratio of a
single trapezoid region, like F, is the ratio of the region in Part C minus the
small triangle in Part A. So, the
ratio is approximately 0.1569. Thus,
the ratio of all three is approximately 3*0.1569=0.4706. Click this GSP link
to investigate.

Thus, the ratio
of the rhombus regions (A, E, G) over the entire triangle is
1-0.4706-.0.01472=approximately 0.5147.
Click this GSP link to investigate.

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