Euclid's WorkshopBook I
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Proposition 30 of 48 • Theorem

Straight lines parallel to the same straight line are also parallel to one another.

If line A is parallel to line C, and line B is also parallel to line C, then A and B are parallel to each other. Parallelism is transitive — it passes through a common parallel.

Before You Read

You know that two lines can each be parallel to a third line—but does that make them parallel to each other? It seems obvious, but obvious things still need proof. Proposition 30 confirms that parallelism is transitive: if A is parallel to C and B is parallel to C, then A is parallel to B. Without this, 'being parallel' would not behave as a coherent geometric relationship.

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// // // A B E F C D G H K

What Euclid Is Doing

Setup: We have three lines: AB, EF, and CD. We are given that AB is parallel to EF, and CD is also parallel to EF. We need to prove that AB is parallel to CD.

Approach: Draw a transversal GHK that crosses all three lines. Use Proposition 29 (properties of parallels) to extract angle equalities from each pair of known parallel lines, then use Proposition 27 to conclude that the third pair is also parallel.

Conclusion: Draw a transversal GHK crossing AB at G, EF at H, and CD at K. Since AB is parallel to EF, the alternate interior angles ∠AGH and ∠GHF are equal (Proposition 29). Since CD is parallel to EF, the alternate interior angles ∠GHF and ∠HKD are equal (Proposition 29 — here ∠GHF is an alternate angle to ∠HKD with respect to transversal HK and parallels EF, CD). So ∠AGH = ∠GHF = ∠HKD, which means ∠AGH = ∠GKD. But ∠AGH and ∠GKD are alternate interior angles formed by the transversal GK crossing lines AB and CD. By Proposition 27, AB is parallel to CD. ✓

Key Moves

  1. Given: AB is parallel to EF, and CD is parallel to EF.
  2. Draw a transversal GHK crossing AB at G, EF at H, and CD at K.
  3. AB parallel to EF implies ∠AGH = ∠GHF (Proposition 29, alternate interior angles).
  4. EF parallel to CD implies ∠GHF = ∠HKD (Proposition 29, alternate interior angles).
  5. Therefore ∠AGH = ∠HKD (Common Notion 1, things equal to the same thing are equal).
  6. ∠AGH and ∠HKD are alternate interior angles for lines AB and CD with transversal GK.
  7. By Proposition 27, AB is parallel to CD ✓

Try It Yourself

Draw three horizontal lines on your page, making the top and bottom ones genuinely parallel (fold the paper to check), then draw the middle one parallel to the bottom one. Are all three mutually parallel? Now try drawing the middle one at a very slight angle away from parallel—does it eventually converge with one of the outer lines even though both outer lines stay parallel to each other?

Proof Challenge

Available Justifications

1.

Draw transversal GK cutting all three lines

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2.

Since AB ∥ EF, alternate angles ∠AGK = ∠GKF

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3.

Since CD ∥ EF, alternate angles ∠GKF = ∠DKG

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4.

∠AGK = ∠DKG (alternate angles equal), so AB ∥ CD

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 30.

  • ✓Lesson Plan (prop-30-lesson-plan.pdf)
  • ✓Student Worksheet (prop-30-worksheet.pdf)
  • ✓Answer Key (prop-30-answer-key.pdf)
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Why It Matters

Transitivity of parallelism is essential for the coherence of the entire parallel theory. Without it, we could not chain parallel relationships: knowing that floor beams are each parallel to a reference line would not tell us they are parallel to each other. This proposition makes the parallel relation well-behaved as an equivalence relation (reflexive, symmetric, transitive).

Modern connection: In coordinate geometry, parallel lines have equal slopes, and equality of real numbers is transitive — so transitivity of parallelism is automatic. But Euclid had no coordinates. He had to prove this structural property from first principles. In modern algebra, this is analogous to proving that an equivalence relation is transitive, which is never trivial without the right axioms.

Historical note: This is only the second proposition in the Elements that depends (indirectly) on the parallel postulate, through its use of Proposition 29. Euclid's proof is elegant in its economy: a single transversal line and two applications of Prop 29 do all the work. The simplicity of the proof belies the depth of the axiom it rests on.

Discussion Questions

  • The proof chains two angle equalities through a common angle at H. This is structurally identical to the transitive property of equality (if a = b and b = c, then a = c). Is this a coincidence, or is there a deeper connection?
  • Does this proposition hold in non-Euclidean geometry? Remember that Proposition 29 fails in hyperbolic geometry. What happens to transitivity of parallelism?
  • Could you prove this proposition without drawing a transversal? What alternative approaches might work?
Euclid's Original Proof
Straight lines parallel to the same straight line are also parallel to one another.

Let each of the straight lines AB, CD be parallel to EF;

I say that AB is also parallel to CD.

For let the straight line GHK fall upon them;

Then, since the straight line GHK has fallen on the parallel straight lines AB, EF,

the angle AGH is equal to the angle GHF. [I.29]

Again, since the straight line GK has fallen on the parallel straight lines EF, CD,

the angle GHF is equal to the angle GKD. [I.29]

But the angle AGH was also proved equal to the angle GHF;

therefore the angle AGH is also equal to the angle GKD; [C.N. 1]

and they are alternate;

therefore AB is parallel to CD. [I.27]

Therefore etc.

Q.E.D.

What's Next

Parallelism is now a fully established, transitive relationship. The next step is to put it to work: Proposition 31 shows how to construct a line through any given point that is parallel to any given line. This construction—drawing a parallel through an external point—is one of the most useful operations in all of geometry, and it depends directly on the angle-copying technique from Proposition 23.