Euclid's WorkshopBook I
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Proposition 39 of 48 • Theorem

Equal triangles which are on the same base and on the same side are also in the same parallels.

If two triangles have equal area, share the same base, and lie on the same side of it, then the line joining their apexes must be parallel to the base. This is the converse of Proposition 37, proved by contradiction.

Before You Read

Proposition 37 said 'same parallels imply equal area.' Now turn it around: if two triangles sharing a base happen to have equal area, are their apexes forced to lie on the same parallel line? It seems obvious, yet proving it requires a proof by contradiction—your first real reductio ad absurdum in the area theory.

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E = area B C A D

What Euclid Is Doing

Setup: Triangles ABC and DBC have equal area, share the same base BC, and lie on the same side of BC. We must prove that AD is parallel to BC — that is, the vertices A and D lie on a line parallel to the base.

Approach: Euclid uses proof by contradiction. Assume AD is not parallel to BC. Then draw a line AE through A that is parallel to BC (Proposition 31), where E is a point different from D. The triangle EBC would then be on the same base BC and between the same parallels as triangle ABC, so by Proposition 37, triangle EBC = triangle ABC. But triangle ABC = triangle DBC (given). So triangle EBC = triangle DBC. But this is impossible because one contains the other (E is not the same point as D, so EBC is either larger or smaller than DBC). This contradiction proves AD must be parallel to BC.

Conclusion: Suppose AD is not parallel to BC. Through A, draw AE parallel to BC (Proposition 31), meeting the line through D at E, where E is not the same point as D. Join EC. Now triangles ABC and EBC are on the same base BC and between the same parallels BC and AE. By Proposition 37, triangle ABC = triangle EBC. But triangle ABC = triangle DBC (given). Therefore triangle EBC = triangle DBC (Common Notion 1). But this is impossible: triangle EBC is a part of triangle DBC (or vice versa), and a part cannot equal the whole (Common Notion 5). This contradiction means our assumption was wrong. Therefore AD is parallel to BC. ✓

Key Moves

  1. Given: triangle ABC = triangle DBC (equal area), same base BC, same side.
  2. Assume for contradiction: AD is NOT parallel to BC.
  3. Through A, draw AE parallel to BC, where E differs from D (Proposition 31).
  4. Triangle ABC and triangle EBC: same base BC, same parallels BC and AE.
  5. By Proposition 37, triangle ABC = triangle EBC.
  6. But triangle ABC = triangle DBC (given), so triangle EBC = triangle DBC.
  7. This is impossible: EBC is part of DBC (or vice versa), contradicting Common Notion 5.
  8. Therefore AD is parallel to BC ✓

Try It Yourself

Draw triangle ABC and then try to draw a second triangle DBC with the same base but a different area—put D slightly above or below the parallel through A. Convince yourself that any such triangle is either bigger or smaller than ABC, making equal area impossible unless D lies exactly on A's parallel.

Proof Challenge

Available Justifications

1.

Given: Triangles ABC and DBC are equal in area and on the same base BC. Join AD.

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2.

Assume AD is NOT parallel to BC. Through A, draw AE parallel to BC (meeting BD extended at E).

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3.

Join EC. Triangles ABC and EBC are on the same base BC and between parallels AE and BC

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4.

Therefore triangle ABC = triangle EBC in area (by Prop 37)

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5.

But triangle ABC = triangle DBC (given). So triangle EBC = triangle DBC.

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6.

But EBC ≠ DBC since E ≠ D (the lesser cannot equal the greater) — contradiction. So AD ∥ BC.

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 39.

  • ✓Lesson Plan (prop-39-lesson-plan.pdf)
  • ✓Student Worksheet (prop-39-worksheet.pdf)
  • ✓Answer Key (prop-39-answer-key.pdf)
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Why It Matters

This is the converse of Proposition 37. Where Proposition 37 says 'same parallels imply equal area,' Proposition 39 says 'equal area implies same parallels.' Together they establish a complete equivalence: for triangles on the same base and same side, equal area is the same thing as lying between the same parallels. This equivalence is essential for the area-based arguments in the Pythagorean theorem.

Modern connection: In analytic geometry, this result says that if two triangles with the same base have equal area, their apexes are at the same perpendicular distance from the base line — they lie on a line parallel to the base. This is the geometric version of the algebraic fact that equal areas with equal bases imply equal heights.

Historical note: This is one of the earliest proofs by contradiction (reductio ad absurdum) in the area theory of the Elements. Euclid uses Common Notion 5 ('the whole is greater than the part') to derive the contradiction — one of the rare occasions where this axiom plays a starring role. The technique of assuming the negation and deriving absurdity became a cornerstone of mathematical reasoning.

Discussion Questions

  • The proof assumes that if E is not the same as D, then one of the triangles EBC, DBC contains the other. Is this always true? What if E is on the opposite side of D from the base?
  • Why does Euclid need proof by contradiction here? Proposition 37 was proved directly. What makes the converse direction harder?
  • Common Notion 5 ('the whole is greater than the part') is crucial to the contradiction. Is this axiom as obvious as it seems? Are there mathematical systems where it fails?
Euclid's Original Proof
Equal triangles which are on the same base and on the same side are also in the same parallels.

Let ABC, DBC be equal triangles which are on the same base BC and on the same side of it;

[I say that they are also in the same parallels.]

And [For] let AD be joined; I say that AD is parallel to BC.

For, if not, let AE be drawn through the point A parallel to the straight line BC, [I.31] and let EC be joined.

Therefore the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels BC, AE. [I.37]

But the triangle ABC is equal to the triangle DBC;

therefore the triangle DBC is also equal to the triangle EBC, [C.N. 1]

the greater to the less: which is impossible.

Therefore AE is not parallel to BC.

Similarly we can prove that neither is any other straight line through A except AD;

therefore AD is parallel to BC.

Therefore etc.

Q.E.D.

What's Next

Proposition 39 proved the converse for the 'same base' case. Proposition 40 closes the circle by proving the same converse result when the bases are merely equal in length rather than identical—completing the four-way symmetry of the area propositions.