You may also like

problem icon

Tri-split

A point P is selected anywhere inside an equilateral triangle. What can you say about the sum of the perpendicular distances from P to the sides of the triangle? Can you prove your conjecture?

problem icon

Fitting In

The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest equilateral triangle which fits into a circle is LMN and PQR is an equilateral triangle with P and Q on the line LM and R on the circumference of the circle. Show that LM = 3PQ

problem icon

No Right Angle Here

Prove that the internal angle bisectors of a triangle will never be perpendicular to each other.

Half a Triangle

Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

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!


The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
Copyright © 1997 - 2012. University of Cambridge. All rights reserved.
NRICH is part of the family of activities in the Millennium Mathematics Project.