Half a Triangle

'Half a Triangle' printed from https://nrich.maths.org/

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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.

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.

$X$ 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!