Proposition 40 of 48 • Theorem

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

If two triangles have equal area, stand on equal-length bases along the same line, and lie on the same side, then the line connecting their apexes is parallel to the base line. This is the converse of Proposition 38, proved by contradiction.

Before You Read

Proposition 39 showed that two equal-area triangles on the same base must have their apexes on the same parallel. Now the bases are separated—does the result still hold when the triangles share only the length of their bases, not the bases themselves?

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What Euclid Is Doing

Setup: Triangles ABC and DEF have equal area, stand on equal bases BC and EF (which lie on the same straight line), and are on the same side of that line. We must prove that AD is parallel to BF (the base line).

Approach: The proof follows the same contradiction pattern as Proposition 39, but uses Proposition 38 (equal bases) instead of Proposition 37 (same base). Assume AD is not parallel to BF. Draw a line through A parallel to BF, meeting a point G different from D. Then triangle GEF would equal triangle ABC by Proposition 38, leading to a contradiction with the given equality of ABC and DEF.

Conclusion: Suppose AD is not parallel to BF. Through A, draw AG parallel to BF (Proposition 31), where G is on the line through D but G is not D. Join GF. Now triangles ABC and GEF are on equal bases BC and EF (given BC = EF) and between the same parallels BF and AG. By Proposition 38, triangle ABC = triangle GEF. But triangle ABC = triangle DEF (given). Therefore triangle GEF = triangle DEF (Common Notion 1). But this is impossible: one of these triangles contains the other (since G and D are different points on the same side), so they cannot be equal (Common Notion 5). This contradiction means AD is parallel to BF. ✓

Key Moves

Given: triangle ABC = triangle DEF (equal area), equal bases BC = EF on the same line, same side.
Assume for contradiction: AD is NOT parallel to BF.
Through A, draw AG parallel to BF, where G differs from D (Proposition 31).
Triangles ABC and GEF: equal bases BC = EF, same parallels BF and AG.
By Proposition 38, triangle ABC = triangle GEF.
But triangle ABC = triangle DEF (given), so triangle GEF = triangle DEF.
This is impossible: GEF is part of DEF (or vice versa), contradicting Common Notion 5.
Therefore AD is parallel to BF ✓

Try It Yourself

Draw two triangles with equal bases on the same line and equal areas but with their apexes seemingly at different heights. Is this actually possible, or does equality of area force the apexes onto the same horizontal? Try to construct a counterexample—you'll find it is impossible, and that attempt is essentially Euclid's proof.

Proof Challenge

Available Justifications

Given: Triangles ABC and DEF are equal in area and on equal bases BC and EF. Join AD.

Drag justification

Assume AD is NOT parallel to BF. Through A, draw AG parallel to BF (meeting DE extended at G).

Drag justification

Join GF. Triangles GBC (with base BC) and DEF (with base EF) are not directly compared — instead, triangle GEF is on base EF between parallels AG and BF

Drag justification

Triangles ABC and GEF are on equal bases BC and EF between parallels AG and BF, so ABC = GEF in area

Drag justification

But triangle ABC = triangle DEF (given). So triangle GEF = triangle DEF.

Drag justification

But GEF ≠ DEF since G ≠ D (the lesser cannot equal the greater) — contradiction. So AD ∥ BF.

Drag justification

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 40.

✓Lesson Plan (prop-40-lesson-plan.pdf)
✓Student Worksheet (prop-40-worksheet.pdf)
✓Answer Key (prop-40-answer-key.pdf)

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Why It Matters

This completes the set of four area propositions and their converses (37/39 and 38/40). Together, they establish a complete equivalence between equal area and equal height for triangles with equal bases. This gives Euclid a powerful tool: to prove two lines are parallel, it suffices to show that certain triangles have equal area — a technique used in the proof of the Pythagorean theorem.

Modern connection: In modern terms, Proposition 40 says that equal area plus equal base implies equal height. This is the geometric version of the algebraic rearrangement: if (1/2)b*h1 = (1/2)b*h2, then h1 = h2. In data visualization and computational geometry, this principle is used when normalizing shapes — triangles with the same base and area must have their apexes at the same distance from the base.

Historical note: Some scholars have questioned whether Proposition 40 is authentically Euclid's or a later interpolation, since its proof is nearly identical to Proposition 39. The manuscript tradition is divided. However, the proposition is logically necessary to complete the parallel-area correspondence for triangles on equal (not just the same) bases, so its inclusion is mathematically justified regardless of authorship.

Discussion Questions

The proof of Proposition 40 is almost identical to Proposition 39, with Proposition 38 replacing Proposition 37. Is there a deeper reason for this structural parallel between the two pairs (35/36 and 37/38)?
Some historians believe Proposition 40 may be a later addition to the Elements. What evidence would you look for to determine whether a proposition is original or interpolated?
Propositions 37-40 give conditions under which triangles have equal area. But they all require the triangles to lie between the same parallels (or prove that they do). Can two triangles with equal area fail to lie between any common pair of parallels?

Euclid's Original Proof

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

Let ABC, CDE be equal triangles on equal bases BC, CE and on the same side.

I say that they are also in the same parallels.

For let AD be joined; I say that AD is parallel to BE.

For, if not, let AF be drawn through A parallel to BE, [I.31] and let FE be joined.

Therefore the triangle ABC is equal to the triangle FCE; for they are on equal bases BC, CE and in the same parallels BE, AF. [I.38]

But the triangle ABC is equal to the triangle DCE;

therefore the triangle DCE is also equal to the triangle FCE, [C.N. 1]

the greater to the less: which is impossible.

Therefore AF is not parallel to BE.

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

therefore AD is parallel to BE.

Therefore etc.

Q.E.D.

What's Next

With propositions 37-40 forming a complete area toolkit, Euclid is now ready for the capstone of Book I's area theory. Proposition 41 establishes that a parallelogram on the same base as a triangle and between the same parallels is exactly double the triangle in area—the precise link that Proposition 47, the Pythagorean theorem, will depend on.