# Half a Triangle

Construct a line parallel to one side of a triangle so that the
triangle is divided into two equal areas.

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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.

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?

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!

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.