Half a Triangle
Construct a line parallel to one side of a triangle so that the
triangle is divided into two equal areas.
Problem
Image
Using only a pair of compasses, and an edge to draw a straight line along,
construct a line parallel to one side of a triangle so that the triangle is divided into two equal areas.
Getting Started
If the area ratio has to be $1:2$ , what would the line ratio need to be?
What constructions do you know that create such a ratio?
Can you see how to create such a construction on the given triangle?
Student Solutions
Well done Arun from National Public School, Bangalore, India, some quality thinking in devising this solution.
We are given a triangle $ABC$ , and are required to draw a line $DE$ parallel to $CB$ such that it divides the triangle into $2$ of equal areas.
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The area of triangle $ABC$ is double the area of $AED$.
But, $ADE$ and $ACB$ are similar triangles because $DE$ is parallel to $CB$We also know that the ratio of the areas of the two similar triangles is equal to the ratio of the squares of corresponding sides.
Which means that the line ratio $AD$:$AC$ must be $1 : \sqrt{2}$
The problem becomes : how to locate $D$ to achieve this
ratio.
A square of side length $1$ has a diagonal length of$
\sqrt{2}$
or, put another way, an isosceles right-angled triangle has a
hypotenuse $ \sqrt{2}$ times bigger than the other sides.
Here is a construction to achieve this required ratio.
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$X$ is any suitable point on $AD$
$ZX$ is perpendicular to $AC$, and $ZX$ is equal in length to $AX$.
So $AXZ$ is an isosceles right-angled triangle.
By sweeping an arc centre $A$ from $X$ to $AZ$ at $N$, $AN$ is made equal to $AX$
$AN$ to $AZ$ is now in the required ratio.
Drawing from $N$ parallel to $ZC$ the point $D$ is reached.
Because $AND$ and $AZC$ are similar triangles, $AD$ and $AC$ are in the required ratio.
Excellent and simple!
Teachers' Resources
This problem will need solving in stages by most students.
Perhaps beginning by establishing the required line ratio.
Then by remembering a familiar figure where that particular ratio can be found.
Deciding how that figure can be constructed on the given triangle to locate a useful point.
Finally creating the required line, parallel to the base, across the triangle.