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Proposition 36 of 48 • Theorem

Parallelograms which are on equal bases and in the same parallels equal one another.

If two parallelograms have equal-length bases (not necessarily the same base) and lie between the same pair of parallel lines, they have equal area. This extends Proposition 35 from 'same base' to 'equal bases.'

Before You Read

Two parallelograms on opposite ends of a long table, nowhere near each other, yet both with the same base length and the same height—must they have the same area? The answer is yes, and the proof is a clever relay race through an intermediate parallelogram.

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What Euclid Is Doing

Setup: We have two parallelograms ABCD and EFGH on equal bases BC and FG, both lying between the same parallel lines. We must prove they have equal area.

Approach: Euclid connects the two parallelograms by drawing lines BE and CH. Since BC = FG (given) and BC is parallel to FG (both lie on the same parallel line), the figure BCFH... actually Euclid uses a clever construction: since BC = FG (equal bases) and both are on the same line, he draws BG and CF. Because BC is equal and parallel to FG (they lie on the same straight line with BC = FG), the joins BG and CF form a new parallelogram BCGF (by Proposition 33). Then both ABCD and EFGH can be compared to this intermediate parallelogram using Proposition 35 (same base).

Conclusion: Join BE and CH. Since BC = FG (given), and both lie on the bottom parallel, while DH lies on the top parallel — Euclid's actual method: join BH and CG. Now BC is equal to FG (given). But FG = EH (opposite sides of parallelogram EFGH, Proposition 34). So BC = EH. Also BC is parallel to EH (both lie between the same parallels). Since BC and EH are equal and parallel, by Proposition 33, BH and CE are also equal and parallel. Therefore BCHE is a parallelogram. Now parallelogram ABCD and parallelogram BCHE are on the same base BC and between the same parallels. By Proposition 35, ABCD = BCHE. Similarly, parallelogram EFGH and parallelogram BCHE are on the same base EH (= FG) — wait, they share base... Euclid's precise argument: EBCH is a parallelogram (just proved). ABCD is on the same base BC as EBCH and between the same parallels, so ABCD = EBCH (Proposition 35). EFGH is on the same base EH as EBCH and between the same parallels, so EFGH = EBCH (Proposition 35). Therefore ABCD = EFGH (Common Notion 1). ✓

Key Moves

  1. Given: parallelograms ABCD and EFGH with equal bases BC = FG, between the same parallels.
  2. BC = FG (given) and FG = EH (opposite sides of EFGH, Proposition 34), so BC = EH (Common Notion 1).
  3. BC is parallel to EH (both on the same parallel line).
  4. Since BC and EH are equal and parallel, join BE and CH: BCHE is a parallelogram (Proposition 33).
  5. ABCD and BCHE share base BC and lie between the same parallels: ABCD = BCHE (Proposition 35).
  6. EFGH and BCHE share base EH and lie between the same parallels: EFGH = BCHE (Proposition 35).
  7. Therefore ABCD = EFGH (Common Notion 1) ✓

Try It Yourself

Draw two clearly separated parallelograms that share no base segment but have equal-length bases and their top sides on the same horizontal line. Sketch the intermediate parallelogram that Euclid constructs to connect them—can you see how Proposition 33 builds that bridge?

Proof Challenge

Available Justifications

1.

Given: Parallelograms ABCD and EFGH on equal bases BC and FG, between parallels AH and BG

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2.

Join BE and CH

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3.

BC = FG (given, equal bases) and EF = HG (opposite sides of parallelogram EFGH), so BC = FG and EF ∥ HG

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4.

BE and CH are equal and parallel (joining equal and parallel lines BC and EH), so EBCH is a parallelogram

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5.

Parallelograms ABCD and EBCH are on the same base BC and between the same parallels, so ABCD = EBCH in area

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6.

Parallelograms EFGH and EBCH are on the same base EH and between the same parallels, so EFGH = EBCH in area

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7.

ABCD = EBCH and EFGH = EBCH, therefore ABCD = EFGH

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 36.

  • ✓Lesson Plan (prop-36-lesson-plan.pdf)
  • ✓Student Worksheet (prop-36-worksheet.pdf)
  • ✓Answer Key (prop-36-answer-key.pdf)
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Why It Matters

This generalizes Proposition 35 from 'same base' to 'equal bases,' removing the requirement that the two parallelograms share a physical base line. Now any two parallelograms with the same base length and height have equal area, no matter where they are positioned. This is the full geometric version of 'area = base times height' for parallelograms.

Modern connection: In coordinate geometry, this becomes the simple algebraic fact that two parallelograms with the same base length and same perpendicular height have the same area. The proposition is also fundamental in the theory of determinants: the absolute value of a 2x2 determinant gives the area of a parallelogram, and two matrices with the same determinant produce parallelograms of equal area.

Historical note: Euclid's proof strategy — introducing an intermediate parallelogram and applying Proposition 35 twice — is an early example of the mathematical technique of 'transitivity through a common comparand.' This elegant argument pattern recurs throughout the Elements.

Discussion Questions

  • The proof introduces a third parallelogram (BCHE) as an intermediary. Why is this intermediate step necessary? Could you prove the result directly from the definition of equal area?
  • Proposition 35 handles the 'same base' case and Proposition 36 handles the 'equal bases' case. Is Proposition 35 just a special case of Proposition 36, or does it carry independent significance?
  • This proposition, combined with Proposition 35, establishes that parallelogram area depends only on base and height. What would it mean geometrically if this were false — if area depended on something else?
Euclid's Original Proof
Parallelograms which are on equal bases and in the same parallels are equal to one another.

Let ABCD, EFGH be parallelograms which are on equal bases BC, FG and in the same parallels AH, BG;

I say that the parallelogram ABCD is equal to EFGH.

For let BE, CH be joined.

Then, since BC is equal to FG while FG is equal to EH, [I.34]

BC is also equal to EH. [C.N. 1]

But they are also parallel.

And EB, HC join them;

but straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are equal and parallel. [I.33]

Therefore EBCH is a parallelogram. [I.33]

And it is equal to ABCD; for it has the same base BC with it, and is in the same parallels BC, AH with it. [I.35]

For the same reason also EFGH is equal to the same EBCH; [I.35]

so that the parallelogram ABCD is also equal to EFGH. [C.N. 1]

Therefore etc.

Q.E.D.

What's Next

Propositions 35 and 36 proved the area results for parallelograms. Euclid now applies the same logic to triangles: Proposition 37 shows that triangles on the same base and between the same parallels are equal in area, linking triangle area directly to the formula half-base-times-height.