*Can you work out how to draw a perfect right angled triangle with two equal sides, using only a straight edge and compass. That is the sort of challenge the Ancient Greek Mathematician Euclid set himself. You need to devise an algorithm to do it – a sequence of steps that guarantee you end up with a right-angled triangle.*

Try it before you read on. It is a fairly simple algorithm. You need to look for a way to draw circles with points that give you a right angle.

Here is my full algorithm for drawing a square. It does give you a square, but it is hard to follow and it was just as hard to work out. I kept getting confused with which circle was which, and being sure what we did really was what I intended…what a nightmare.

- Draw a circle with radius AB centred on A, a point on a line to give a point B where line and circle cross.
- Draw a circle with radius AB centred on point B. This gives two points C and D where the circles cross.
- Draw a line through C and D. This is now at right angles to the original line. This gives a new point E where the two lines cross.
- Draw a circle with radius EC, centred on E. This gives a new point F where the circle crosses the original line in the circle.
- Draw a line CF.
- You have a perfect right angled triangle with two equal sides: CEF.

Given a right-angled triangle like this can you now see how to draw a square, starting with it (again only using a compass and straight edge).

Can you see how we could have decomposed this into two steps? (1. draw a line, perpendicular to another line 2. turn it into a triangle with 2 equal sides)

Can you work out how to construct a right-angled triangle but where its right-angle corner must be at a given point that you start with?

Can you prove (i.e. provide a totally convincing logical argument) that the algorithm works? i.e. that it is a right-angled triangle with two equal sides?

More on Euclid and algorithms for creating perfect shapes

This work was supported by the Institute of Coding, which is supported by the Office for Students (OfS).

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