In this post, we will show you how to construct an equilateral triangle, when you are given nothing but a line segment to start. In the process, we will construct the same diagram found in Euclid’s The Elements, Book I, Proposition 1. This is also the same diagram used in our earlier post “Proving A Triangle Is Equilateral”.

Step 1: We use a straightedge to create a line segment. We name one endpoint A, and the other endpoint B. Not surprisingly, we name this line segment AB.

Step 2: We use our compass to construct a circle around A. We push the pin of the compass into point A, and adjust the compass until the pencil lies on point B. In doing this, we have set the compass to construct a circle with a radius equal to the length of AB. While keeping the pin of the compass pressed into point A, we gently spin the compass around. As the pencil drags across the paper, it draws out our first circle! We have now constructed a circle centered at point A, with radius AB.

Step 3: We then pick up the compass, and push the pin into point B. Line up the pencil of the compass with point A; it should already be the correct radius. We will now use the compass to draw another circle, this one centered at B and with BA as a radius. When we are finished, we will have two identical, overlapping circles who centers are connected by the line segment AB.

Step 4: See that point above line segment AB where the two circles intersect? Let’s call that point C.

Step 5: Now, we’ll draw a line segment from point A to point C. Since both AB and AC are radii of circle A, they must be the same length. We’ll put a small tick mark on each one to remind us of this congruency.

Step 6: Finally, we will draw our last line segment, from point B to point C. Since both BA and BC are radii of circle B, they must be the same length. We’ll put a small double tick mark on each one to remind us of this congruency.

Step 7: At this point, we have now constructed a triangle. As we have proven earlier, this triangle ABC must be equilateral. I’ve outlined the triangle in bright blue, marked each side with triple tick marks, and lightly shaded the inside of the triangle.

Congratulations! You have now constructed a perfectly equilateral triangle, without ever having to use a ruler to measure length, or a protractor to measure angles.

Paper, Pencil, Straightedge, and Compass.

To do geometric constructions, you need four items: paper, pencil, a straightedge, and a compass. For a straightedge, any good ruler will suffice. You can buy a cheap compass at your local Walgreen’s, Rite-Aid, or whatever for about $3.

Don’t worry about whether the ruler is marked in inches or centimeters. Similarly, don’t panic if your compass has absolutely no markings at all! Many compasses used for drafting don’t; the manufacturers figure you’ll line the compass up against a ruler and set the radius that way. As you’ll soon discover, however, you don’t need a ruler to find the midpoint of a straight line; in a later post we’ll show you how a compass can do that very nicely…

Let’s take a look at the diagram below (figure 1) discussed in The Elements, Book I, Proposition 1.

First, take a look at line segment AB. You’ll notice that AB is a radius to two circles. The circle on the left is centered at point A, and has AB as a radius. On the other hand, the circle on the right is centered at point B, and has BA as a radius.

Above line segment AB lies point C. You’ll notice that point C is one of two points where circles A and B intersect. We also have a line segment from A to C, and another one from B to C. We have a neat little triangle ABC in the intersection of the two circles.

What kind of triangle is it? It sure looks like it could be an equilateral triangle.

But just saying that it looks like an equilateral triangle doesn’t mean that it actually is an equilateral triangle. We must find a way to prove it.

Here’s what we know for certain, what we are given:

1) Circle A has radius AB

2) Circle B has radius BA

3) Circles A and B intersect at point C

From these three statements, we will now prove that triangle ABC is equilateral.

Step 1: Since C is the intersection point between circles A and B, point C lies on both circle A and circle B.

Step 2: Therefore, line segment AC is a radius of circle A.

Step 3: Since all radii (plural of radius, pronounced RAY-DEE-EYE) of a given circle are equal, AB=AC.

So far, all we’ve done is use a simple property of circles to prove that line segments AB and AC are congruent. Now, we will use the same reasoning on circle B.

Step 4: By Step 1, line segment BC is a radius of circle B.

Step 5: By step 3, BA=BC.

At this point, we’ve proven that AB=AC, and that BA=BC. This may seem ridiculously obvious, but it’s also worth pointing out that AB=BA. What we have here is simply two different names for the same line segment. Traditionally, when we name the radius of a circle, we let the first letter be the center of the circle, and the second letter be a point on the circle. AB, therefore, is the radius of circle A, where A is the center of the circle, and B is a point on the circle. Similarly, BA is the radius of circle B, where B is the center of the circle, and A is a point on the circle. In the end, however, we simply have two names for the same object.

As a result of this, we can rewrite BA=BC as AB=BC.

Step 6: Since AB=AC, and AB=BC, it then follows that AC=BC.

You may remember this statement from your algebra classes as the “transitive property”, which says that if a=b, and b=c, then a=b=c. In the original work The Elements by Euclid, this statement is called “Common Notion #1″. The generally accepted English translation (remember, Euclid lived 2200 years ago in ancient Greece) of Common Notion #1 is “Things which are equal to the same thing are also equal to one another.” Whether you want to refer to it as Common Notion #1 or the transitive property, the resulting claim we can make is the same:

Step 7: Since AB=AC=BC, all three sides of triangle ABC are equal

Finally, we use this statement to show that triangle ABC fits the definition of an equilateral triangle.

Step 8: Since all three sides of triangle ABC are equal, triangle ABC is equilateral.

End of proof.

We could also rewrite this proof into something resembling a normal English paragraph:

“Let there be two circles A and B, whose centers are connected by the line segment AB, such that AB is a radius for both circles. Let C be a point of intersection between circles A and B. Since AB and AC are both radii of circle A, AB=AC. By similar reasoning, AB=BC. Since AB=AC=AB, triangle ABC is equilateral. End of proof.”

Take a look at where the proof has been streamlined. No mention is made of Common Notion #1 or the transitive property; the proof assumes that the reader already knows it. This is, in fact, a very safe assumption for any reader who has had math through Algebra II. The proof also assumes that the reader knows that all radii of a given circle are equal. A very safe assumption, again, for any reader who has had one full semester of high school geometry.

Writing proofs can be a tricky business. In general, when first starting out, you want to err on the side of too many steps rather than too few. As your mastery of the material grows, you will find yourself streamlining your own proofs automatically, often without realizing it. Make sure that you yourself can follow your own proof! If you can’t follow it yourself, then how do you expect someone else to follow it?