Euclid's WorkshopBook I
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Proposition 22 of 48 • Construction

To construct a triangle out of three straight lines which equal three given straight lines: thus it is necessary that the sum of any two of the straight lines should be greater than the remaining one.

Given three line segments (where any two together are longer than the third), build a triangle whose sides are exactly those three lengths.

Before You Read

You are handed three sticks. Can you always form a triangle from them? Not necessarily—if one stick is longer than the other two combined, no triangle is possible. But if every stick is shorter than the sum of the other two, a triangle must exist. Can you construct one using only compass and straightedge, with no measuring allowed?

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a b c A B C D F G H E K L

What Euclid Is Doing

Setup: We are given three line segments a, b, and c, with the condition that any two of them sum to more than the third (the triangle inequality). We must construct a triangle with sides equal to these three segments.

Approach: Euclid sets out a working line DE and lays the three lengths end to end along it: DF = a, FG = b, GH = c (Proposition 3). He then draws a circle centered at F with radius FD and a circle centered at G with radius GH — each radius is anchored at its own center, so Postulate 3 applies directly. The circles meet at K, and triangle KFG has sides KF = a, FG = b, GK = c.

Conclusion: Set out a line DE and cut off DF = a, FG = b, GH = c laid end to end (Proposition 3). Draw the circle centered at F with radius FD (Postulate 3) and the circle centered at G with radius GH (Postulate 3). The triangle inequality guarantees these circles intersect—let K be an intersection point. Join KF and KG (Postulate 1). Now KF = FD = a (radii, Definition 15), FG = b by construction, and GK = GH = c (radii, Definition 15). Triangle KFG has sides a, b, c as required. ✓

Key Moves

  1. Set out a line DE and lay the three lengths along it: DF = a, FG = b, GH = c (Proposition 3)
  2. Draw a circle centered at F with radius FD (Postulate 3) — the radius is anchored at its center
  3. Draw a circle centered at G with radius GH (Postulate 3)
  4. The triangle inequality (Proposition 20) ensures the circles intersect. Let K be an intersection point.
  5. Join KF and KG (Postulate 1). KF = FD = a and GK = GH = c by Definition 15 (radii).
  6. Triangle KFG has sides KF = a, FG = b, GK = c ✓

Try It Yourself

Draw three line segments of lengths roughly 5 cm, 7 cm, and 9 cm. Using Proposition 3 to cut off segments and Postulate 3 to draw circles, try building Euclid's construction: lay down one segment, then use two circles centered at its endpoints to locate the third vertex. Does the point of intersection always appear exactly where you expect?

Proof Challenge

Available Justifications

1.

Set out a line DE, and on it cut off DF = a, then FG = b, then GH = c, laid end to end

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2.

Draw the circle with center F and radius FD

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3.

Draw the circle with center G and radius GH

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4.

Let the circles meet at K; join KF and KG

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5.

KF = FD = a and KG = GH = c (radii of the two circles), and FG = b by construction — triangle KFG has the three required side lengths

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0 of 5 steps completed

Curriculum Materials

Get the Teaching Materials

The lesson plan, student worksheet, and answer key for Proposition 22 come with the curriculum bundles.

  • ✓Included in Foundations (Propositions 1–26)
  • ✓or the Complete Collection (all 48)
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Why It Matters

This is the triangle existence theorem: it tells you exactly when three lengths can form a triangle and shows how to build it. It is the constructive companion to the triangle inequality (Prop 20), which gave the necessary condition. Together they completely characterize which triples of lengths are 'triangulable.'

Going deeper

Modern connection: This construction is the basis for triangulation in surveying and GPS. Given three measured distances, you construct a triangle to locate a point. In computer graphics, mesh generation algorithms repeatedly solve this problem when building 3D surfaces from edge lengths.

Historical note: This is one of the few propositions where Euclid explicitly states a necessary condition (the triangle inequality) as part of the problem statement. Usually he leaves such conditions implicit. The explicit mention suggests he recognized the importance of the constraint and wanted to highlight that construction is impossible without it.

Discussion Questions

  • Why does Euclid need the triangle inequality condition stated upfront? What goes wrong in the construction if it fails?
  • The two circles could intersect at two points. Does the choice of intersection point matter? How many different triangles can you get?
  • How does this construction relate to the SSS congruence theorem (Proposition 8)? If you construct two triangles from the same three lengths, must they be congruent?
Euclid's Original Proof
Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. [I.20]

Let the three given straight lines be A, B, C, and of these let two taken together in any manner be greater than the remaining one, namely A, B greater than C; A, C greater than B; and B, C greater than A;

thus it is required to construct a triangle out of straight lines equal to A, B, C.

Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E, and let DF be made equal to A, FG equal to B, and GH equal to C. [I.3]

With centre F and distance FD let the circle DKL be described; [Post. 3]

again, with centre G and distance GH let the circle KLH be described; [Post. 3]

and let KF, KG be joined; [Post. 1]

I say that the triangle KFG has been constructed out of three straight lines equal to A, B, C.

For, since the point F is the centre of the circle DKL, FD is equal to FK. [Def. 15]

But FD is equal to A;

therefore KF is also equal to A. [C.N. 1]

Again, since the point G is the centre of the circle LKH, GH is equal to GK. [Def. 15]

But GH is equal to C;

therefore KG is also equal to C. [C.N. 1]

And FG is also equal to B;

therefore the three straight lines KF, FG, GK are equal to the three straight lines A, B, C.

Therefore out of the three straight lines KF, FG, GK, which are equal to the three given straight lines A, B, C, the triangle KFG has been constructed.

(Being) what it was required to do.

What's Next

Proposition 22 lets us build a triangle from three given lengths. But what if what you need to copy is not a triangle but a single angle? Transferring an angle to a new location is the next fundamental construction, and Proposition 23 solves it by cleverly reducing the problem to the triangle-construction you just learned.