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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Perpendiculars & Angles (5)

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Interior Triangles & Angle Copying (5)

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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 links the two parallelograms through an intermediate one. The segments BC and EH are equal (BC = FG is given, and FG = EH by Proposition 34) and lie on the two parallels, so joining BE and CH produces a parallelogram EBCH (Proposition 33). Each given parallelogram shares a base with EBCH — ABCD shares BC, and EFGH shares EH — so each equals EBCH by Proposition 35, and hence they equal each other.

Conclusion: Join BE and CH (Postulate 1). BC = FG (given), and FG = EH (opposite sides of parallelogram EFGH, Proposition 34), so BC = EH (Common Notion 1). BC is also parallel to EH (they lie on the two parallels). Since BE and CH join the equal and parallel segments BC and EH at their endpoints, they are themselves equal and parallel (Proposition 33) — so EBCH is a parallelogram. ABCD and EBCH stand on the same base BC and between the same parallels, so ABCD = EBCH (Proposition 35). EFGH and EBCH stand on the same base EH and between the same parallels, so EFGH = EBCH (Proposition 35). Therefore ABCD = EFGH (Common Notion 1). ✓

Key Moves

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

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

Drag justification

Join BE and CH

Drag justification

BC = FG (given) and FG = EH (opposite sides of parallelogram EFGH), so BC = EH

Drag justification

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

Drag justification

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

Drag justification

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

Drag justification

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

Drag justification

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Curriculum Materials

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The lesson plan, student worksheet, and answer key for Proposition 36 come with the curriculum bundles.

✓Included in Advanced (Propositions 27–48)
✓or the Complete Collection (all 48)

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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.

Going deeper

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.