Euclid's WorkshopBook I
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Proposition 1 of 48 • Construction

To construct an equilateral triangle on a given finite straight line.

Given any line segment, build a triangle where all three sides are the same length, using that segment as one of the sides.

Before You Read

Grab a compass and straightedge (or just imagine them). Someone hands you a line segment. Can you build a triangle where all three sides are exactly the same length? No rulers, no measuring—just the compass and straightedge. Try it before reading on.

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Definitions (23)
Postulates (5)
Common Notions (5)
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Basic Constructions (5)
Triangle Fundamentals (5)
Perpendiculars & Angles (5)
Exterior Angles & Inequalities (5)
Interior Triangles & Angle Copying (5)
Parallel Lines (5)
Parallel Constructions & Parallelograms (5)
Area Theorems (5)
Area Applications (4)
Grand Finale (4)
A B C D E

What Euclid Is Doing

Setup: We're given a line segment AB. Our goal: construct an equilateral triangle using AB as one side.

Approach: Euclid's strategy is elegant. He draws two circles—one centered at A with radius AB, one centered at B with radius BA. Where they intersect gives us point C, which is the same distance from both A and B.

Conclusion: Here's why all three sides are equal: AC = AB because both are radii of the circle centered at A (Definition 15: all radii of a circle are equal). BC = BA because both are radii of the circle centered at B (Definition 15). AB = BA because they're the same segment. Now we chain: AC = AB (radii of A-circle), BC = AB (radii of B-circle, and BA = AB). By Common Notion 1 (things equal to the same thing are equal to each other): AC = AB = BC. All three sides are equal, so triangle ABC is equilateral by Definition 20. ✓

Key Moves

  1. Draw a circle centered at A with radius AB (Postulate 3). Any point on this circle is distance AB from A.
  2. Draw a circle centered at B with radius BA (Postulate 3). Any point on this circle is distance BA from B.
  3. Mark point C where the two circles intersect. C lies on both circles.
  4. Draw lines AC and BC (Postulate 1)
  5. Since C is on the A-circle: AC = AB (Definition 15—radii are equal)
  6. Since C is on the B-circle: BC = BA = AB (Definition 15)
  7. Therefore AC = AB = BC (Common Notion 1), and triangle ABC is equilateral ✓

Try It Yourself

Draw a line segment of any length on paper. Now perform the construction: two circles, find the intersection, connect the dots. Does it work no matter how long or short your starting segment is? Try a very short segment and a very long one.

Proof Challenge

Available Justifications

1.

Draw a circle centered at A passing through B

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2.

Draw a circle centered at B passing through A

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3.

Mark point C where the circles intersect. Draw lines AC and BC.

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4.

AC = AB (both are radii of the circle centered at A)

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5.

BC = BA (both are radii of the circle centered at B)

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6.

AC = AB and BC = AB, therefore AC = BC

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7.

Since AC = AB = BC, triangle ABC is equilateral

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0 of 7 steps completed

Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 1.

  • ✓Lesson Plan (prop-01-lesson-plan.pdf)
  • ✓Student Worksheet (prop-01-worksheet.pdf)
  • ✓Answer Key (prop-01-answer-key.pdf)
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Why It Matters

This is Proposition 1 for a reason—it's the foundation. Euclid immediately shows that the tools we have (straightedge and compass) can create something useful. The equilateral triangle becomes a building block for later constructions.

Modern connection: Equilateral triangles appear everywhere: in engineering (truss structures), architecture (geodesic domes), nature (honeycombs), and design (the recycling symbol). Their perfect symmetry makes them incredibly useful.

Historical note: There's actually a small logical gap here that wasn't noticed for centuries: Euclid assumes the two circles intersect. He doesn't prove it. This seems obvious visually, but pure logic requires it to be stated.

Discussion Questions

  • Why do you think Euclid chose this as Proposition 1? What makes it a good starting point?
  • Could you construct an equilateral triangle with just a straightedge (no compass)? Why or why not?
  • The two circles intersect at two points. Why does it not matter which one we pick for C?
Euclid's Original Proof
Let AB be the given finite straight line.

Thus it is required to construct an equilateral triangle on the straight line AB.

With centre A and distance AB let the circle BCD be described; [Post. 3]
again, with centre B and distance BA let the circle ACE be described; [Post. 3]
and from the point C, in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined. [Post. 1]

Now, since the point A is the centre of the circle CDB, AC is equal to AB. [Def. 15]

Again, since the point B is the centre of the circle CAE, BC is equal to BA. [Def. 15]

But CA was also proved equal to AB; therefore each of the straight lines CA, CB is equal to AB.

And things which are equal to the same thing are also equal to one another; [C.N. 1]
therefore CA is also equal to CB.

Therefore the three straight lines CA, AB, BC are equal to one another.

Therefore the triangle ABC is equilateral; and it has been constructed on the given finite straight line AB.

(Being) what it was required to do.

What's Next

We can now build an equilateral triangle anywhere we want. But what if we need to copy a specific length to a completely different location? That turns out to be surprisingly tricky with just a compass and straightedge—and it's exactly what Proposition 2 tackles.