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Proposition 35 of 48 • Theorem

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

If two parallelograms share the same base and their opposite sides lie on the same parallel line, then they have equal area — even if they look completely different in shape. This is Euclid's first theorem about area equality without congruence.

Before You Read

Picture a parallelogram like a stack of cards: slide the top layer sideways and the shape changes dramatically, yet every card still covers the same width. Could the area truly stay the same no matter how far you slide?

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// // A B C D E F G base BC

What Euclid Is Doing

Setup: We have two parallelograms ABCD and EBCF on the same base BC, between the same parallels AF and BC (the tops AD and EF both lie along AF; the base BC lies on the other parallel). We must prove they have equal area.

Approach: Euclid's strategy is a cut-and-paste argument. Both top sides AD and EF lie on the same parallel, and each equals the base BC (Proposition 34), so AD = EF. Adding the common piece DE shows AE = DF, which makes the two slanted triangles EAB and FDC congruent by SAS. Subtracting the small overlap triangle DGE from each, and then adding back the shared triangle GBC, turns one parallelogram into the other — so their areas are equal.

Conclusion: AD = BC and EF = BC (opposite sides of the parallelograms, Prop 34), so AD = EF (CN1). Add the common segment DE: AE = DF (CN2). Also AB = DC (Prop 34), and ∠EAB = ∠FDC (corresponding angles on the parallels, Prop 29). So triangles EAB and FDC are congruent by SAS (Prop 4). Subtract the common triangle DGE from each: trapezoid ABGD = trapezoid EGCF (CN3). Add the common triangle GBC to each: parallelogram ABCD = parallelogram EBCF. ✓

Key Moves

  1. Given: parallelograms ABCD and EBCF on the same base BC, between the parallels AF and BC.
  2. AD = BC and EF = BC (opposite sides, Proposition 34), so AD = EF (Common Notion 1).
  3. Add the common segment DE to both: AE = DF (Common Notion 2).
  4. AB = DC (opposite sides of ABCD, Proposition 34), and ∠EAB = ∠FDC (corresponding angles, Proposition 29).
  5. Triangles EAB and FDC are congruent by SAS (Proposition 4).
  6. Subtract the common triangle DGE from each: trapezoid ABGD = trapezoid EGCF (Common Notion 3).
  7. Add the common triangle GBC to each: parallelogram ABCD = parallelogram EBCF ✓

Try It Yourself

On graph paper, draw two parallelograms that share the base from (0,0) to (5,0) but have their top sides on the line y = 3—one with vertices at (1,3) and (6,3), the other wildly slanted with vertices at (−2,3) and (3,3). Count the grid squares inside each. Are the areas equal?

Proof Challenge

Available Justifications

1.

Given: Parallelograms ABCD and EBCF on the same base BC, between parallels AF and BC

Drag justification
2.

AD = BC, EF = BC, and AB = DC (opposite sides of the parallelograms)

Drag justification
3.

AD = EF, since both equal BC

Drag justification
4.

AD = EF and DE is common, so AD + DE = EF + DE, i.e. AE = DF

Drag justification
5.

∠FDC = ∠EAB (corresponding angles, since AB ∥ DC with the transversal along AF)

Drag justification
6.

In triangles EAB and FDC: AE = DF, AB = DC, ∠EAB = ∠FDC. So △EAB ≅ △FDC by SAS.

Drag justification
7.

Subtract triangle DGE from both congruent triangles; the remaining areas are equal: trapezoid ABGD = trapezoid EGCF

Drag justification
8.

Add triangle GBC to both: ABCD = EBCF — the parallelograms have equal area

Drag justification
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Curriculum Materials

Get the Teaching Materials

The lesson plan, student worksheet, and answer key for Proposition 35 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 is Euclid's first area theorem that goes beyond simple congruence. Two parallelograms can look completely different — one tall and narrow, the other wide and slanted — yet if they share the same base and height (between the same parallels), their areas are equal. This is the geometric origin of the formula 'area = base times height.' It opens the entire area theory that culminates in the Pythagorean theorem (Proposition 47).

Going deeper

Modern connection: The formula 'area of a parallelogram = base times height' is taught in every school. Proposition 35 is the geometric version of this formula, proved without any notion of numerical measurement. In calculus, this same principle — that area depends only on base and height, not on the slant — reappears in Cavalieri's principle, which is the foundation of integration.

Historical note: This proposition marks a conceptual leap in the Elements. Until now, area equality meant congruence — same shape, same size. Here, for the first time, Euclid proves that differently shaped figures can have equal areas. This distinction between 'equal' (in area) and 'congruent' (identical in shape) was one of Euclid's most sophisticated ideas, and it confused commentators for centuries.

Discussion Questions

  • Two parallelograms with the same base and between the same parallels have equal area, even though they may look completely different. How does this relate to the modern formula 'area = base times height'?
  • The proof involves adding and subtracting regions (Common Notions 2 and 3). Euclid treats area like a quantity that obeys arithmetic rules. Is this justified? What assumptions is Euclid making about area?
  • This is the first proposition where 'equal' means 'equal in area' rather than 'congruent.' Why is this distinction important, and how does it change the nature of geometric reasoning from this point forward?
Euclid's Original Proof
Parallelograms which are on the same base and in the same parallels are equal to one another.

Let ABCD, EBCF be parallelograms on the same base BC and in the same parallels AF, BC;

I say that ABCD is equal to the parallelogram EBCF.

For, since ABCD is a parallelogram, AD is equal to BC. [I.34]

For the same reason also EF is equal to BC,

so that AD is also equal to EF; [C.N. 1]

and DE is common;

therefore the whole AE is equal to the whole DF. [C.N. 2]

But AB is also equal to DC; [I.34]

therefore the two sides EA, AB are equal to the two sides FD, DC respectively,

and the angle FDC is equal to the angle EAB, the exterior to the interior; [I.29]

therefore the base EB is equal to the base FC,

and the triangle EAB will be equal to the triangle FDC. [I.4]

Let DGE be subtracted from each;

therefore the trapezium ABGD which remains is equal to the trapezium EGCF which remains. [C.N. 3]

Let the triangle GBC be added to each;

therefore the whole parallelogram ABCD is equal to the whole parallelogram EBCF. [C.N. 2]

Therefore etc.

Q.E.D.

What's Next

Proposition 35 handles the case where two parallelograms share the same base. Proposition 36 immediately generalizes this: equal-length bases anywhere along the same parallel are enough—the parallelograms don't even need to touch.