Construct a segment parallel to a base of a triangle that divides the triangle into two equal areas.



The segment that we want to construct will look something like this:

The segment MN divides triangle ABC into two equal areas, so the area of triangle MNC is equal to the area of trapezoid ABNM.


How do we construct the desired segment MN?

Since the desired segment MN divides triangle ABC into two parts of equal area, we know that the area of triangle MNC must be half the area of triangle ABC. Since the ratio of their areas is 1/2, the ratio of their side lengths must be 1/sqrt(2).

Thus, if the height of triangle MNC is h and the height of triangle ABC is H, then we know that

H = h * sqrt (2).

We know H, so now we need to construct a segment having length H / sqrt (2).  If we construct a right triangle with hypotenuse of length H, the legs of the triangle will have length H / sqrt(2), or h.  (We can construct a right triangle with a given hypotenuse by using this script.)

Now that we have a segment of length h, we need to construct the altitude of triangle MNC.  Since segment MN must be parallel to segment AB, the altitude of MNC must be perpendicular to AB.  In other words, it is part of the segment H.  We can construct a circle with center C and radius h.  The intersection of this circle the segment H forms the foot of the altitude.

We can construct a line through this point of intersection that a parallel to segment AB.  The points of intersection of this line and the sides AC and AB give us the points M and N, respectively.


Since the area of triangle MNC is half the area of triangle ABC, triangle MNC and trapezoid ABNM have equal areas.


To view a GSP sketch of the construction, click here.

For a GSP 3 script of the construction, click here.


To go to Part 2 of the problem, click here.



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