Proposition 2 of 48 • Construction

To place a straight line equal to a given straight line with one end at a given point.

Given a line segment and a point somewhere else, construct a new line segment starting at that point that's exactly the same length as the given one.

Before You Read

Here's a puzzle: you have a line segment BC and a point A somewhere else on the page. You need to create a new line starting at A that's exactly the same length as BC. The catch? Your compass 'collapses' when you lift it—you can't just measure BC and redraw it at A. How would you do it?

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What Euclid Is Doing

Setup: We have a line segment BC and a point A located somewhere else. We want to create a new line starting at A that has the exact same length as BC. But look carefully at our postulates: Postulate 3 lets us draw a circle centered at a point passing through another point. It does NOT directly let us draw a circle 'with the same radius as BC' centered somewhere else—that would require the radius to exist as a transferable quantity independent of the points that define it.

Approach: The key insight is using the equilateral triangle as a 'relay station.' Since DA = DB (equilateral triangle), if we can get the length BC onto the line through D and B, we can transfer that same length onto the line through D and A.

Conclusion: Here's why AL = BC: The circle at B with radius BC gives us BG = BC. The circle at D with radius DG gives us DL = DG. Since G is on DB extended, DG = DB + BG. Since L is on DA extended, DL = DA + AL. Now: DG = DL (both radii of the D-circle), and DA = DB (equilateral triangle). Therefore: DB + BG = DA + AL → DB + BC = DB + AL → BC = AL. ✓

Key Moves

Connect A to B with a straight line (Postulate 1)
Construct equilateral triangle DAB on line AB (Proposition 1)—this gives us DA = DB = AB
Extend DA beyond A, and extend DB beyond B (Postulate 2)
Draw circle centered at B with radius BC (Postulate 3)—mark G where it crosses DB extended. Now BG = BC.
Draw circle centered at D with radius DG (Postulate 3)—mark L where it crosses DA extended. Now DL = DG.
Since DG = DB + BG and DL = DA + AL, and DA = DB, we get AL = BG = BC

Try It Yourself

Draw a short segment BC and a point A far away from it. Now try to copy that length to A using only the compass and straightedge. Hint: you'll need Proposition 1's equilateral triangle as a stepping stone. It takes more steps than you'd expect.

Proof Challenge

Available Justifications

Connect A to B with a straight line

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Construct equilateral triangle DAB on line AB, giving DA = DB = AB

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Extend DA beyond A, and extend DB beyond B

Drag justification

Draw circle centered at B with radius BC — mark G where it crosses DB extended.

Drag justification

Because G lies on the circle centered at B, BG = BC.

Drag justification

Draw circle centered at D with radius DG — mark L where it crosses DA extended.

Drag justification

Because L lies on the circle centered at D, DL = DG.

Drag justification

DG = DB + BG and DL = DA + AL. Since DA = DB, we get AL = BG = BC.

Drag justification

0 of 8 steps completed

Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 2.

✓Lesson Plan (prop-02-lesson-plan.pdf)
✓Student Worksheet (prop-02-worksheet.pdf)
✓Answer Key (prop-02-answer-key.pdf)

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Why It Matters

This proposition proves that 'copying a length to a new location' can be derived from Euclid's basic postulates. Postulate 3 only permits drawing a circle through a specific point—it doesn't directly allow transferring a radius. By proving this construction works, Euclid shows his minimal set of postulates is sufficient for all the length operations we might want.

Modern connection: This demonstrates Euclid's axiomatic rigor. He doesn't assume any operation is valid just because it seems obvious—he proves it from first principles. This same approach underlies all of modern mathematics: start with minimal axioms, derive everything else.

Historical note: This is one of the most involved constructions in Book I for such a simple result. It demonstrates Euclid's commitment to proving everything from minimal assumptions. The construction shows that his postulates are 'complete' in a practical sense.

Discussion Questions

Look at Postulate 3: what operation does it actually permit? Why doesn't it directly allow copying a length?
How does the equilateral triangle help us transfer the length?
What does this proposition tell us about the power of Euclid's postulates?

Euclid's Original Proof

Let A be the given point, and BC the given straight line.

Thus it is required to place at the point A as an extremity a straight line equal to the given straight line BC.

From the point A to the point B let the straight line AB be joined; [Post. 1]
and on it let the equilateral triangle DAB be constructed. [I. 1]

Let the straight lines AE, BF be produced in a straight line with DA, DB; [Post. 2]
with centre B and distance BC let the circle CGH be described; [Post. 3]
and again, with centre D and distance DG let the circle GKL be described. [Post. 3]

Then, since the point B is the centre of the circle CGH, BC is equal to BG.

Again, since the point D is the centre of the circle GKL, DL is equal to DG.

And in these DA is equal to DB; therefore the remainder AL is equal to the remainder BG. [C.N. 3]

But BC was also proved equal to BG; therefore each of the straight lines AL, BC is equal to BG.

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

Therefore at the given point A the straight line AL is placed equal to the given straight line BC.

(Being) what it was required to do.

What's Next

Now we can copy any length to any point. The natural next question: can we trim a longer line to match a shorter one? That's Proposition 3—and it's much simpler now that we have Proposition 2 in our toolkit.