Euclid's WorkshopBook I
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Proposition 37 of 48 • Theorem

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

If two triangles share the same base and their opposite vertices lie on a line parallel to the base, then the two triangles have equal area — regardless of where on the parallel line those vertices sit.

Before You Read

Take a triangle and slide its apex left or right along a line parallel to the base—the shape shifts dramatically but something remains constant. What is staying the same, and why?

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// // B C A D E F base

What Euclid Is Doing

Setup: Triangles ABC and DBC are on the same base BC and between the same parallel lines. Vertex A and vertex D both lie on the line parallel to BC. We must prove the triangles have equal area.

Approach: Euclid's strategy is to construct a parallelogram around each triangle, then use the fact that each triangle is half its parallelogram (Proposition 34) and that the two parallelograms are equal (Proposition 35). Through point A, draw a line parallel to BC's direction to form parallelogram EBCA. Through point D, draw a line to form parallelogram DBCF. Both parallelograms share base BC and lie between the same parallels, so they are equal in area (Proposition 35). Each triangle is half its respective parallelogram (Proposition 34), so the triangles are equal.

Conclusion: Through A, draw AE parallel to BC (Proposition 31). Through D, draw DF parallel to BC (Proposition 31). Wait — we need parallelograms on base BC. Through B, draw BE parallel to CA (Proposition 31), meeting the upper parallel at E. Through C, draw CF parallel to BD (Proposition 31), meeting the upper parallel at F. Now EBCA is a parallelogram (BE parallel to CA, and EA parallel to BC). DBCF is a parallelogram (BD parallel to CF, and DF parallel to BC). Both parallelograms are on the same base BC and between the same parallels. By Proposition 35, parallelogram EBCA = parallelogram DBCF. But the diagonal AC bisects parallelogram EBCA (Proposition 34), so triangle ABC = half of EBCA. And the diagonal BD bisects parallelogram DBCF (Proposition 34), so triangle DBC = half of DBCF. Halves of equal things are equal (Common Notion 3 — if equals are subtracted from equals, the remainders are equal). Therefore triangle ABC = triangle DBC. ✓

Key Moves

  1. Given: triangles ABC and DBC on the same base BC, between the same parallels.
  2. Through B, draw BE parallel to CA; EBCA is a parallelogram (Proposition 31).
  3. Through C, draw CF parallel to BD; DBCF is a parallelogram (Proposition 31).
  4. Parallelograms EBCA and DBCF share base BC and lie between the same parallels.
  5. By Proposition 35, parallelogram EBCA = parallelogram DBCF.
  6. Diagonal AC bisects EBCA: triangle ABC = half of EBCA (Proposition 34).
  7. Diagonal BD bisects DBCF: triangle DBC = half of DBCF (Proposition 34).
  8. Halves of equals are equal: triangle ABC = triangle DBC (Common Notion 3) ✓

Try It Yourself

Draw a horizontal base BC and a horizontal line above it. Place a point A anywhere on the upper line and draw triangle ABC, then place point D somewhere else on the same upper line and draw triangle DBC. Calculate or estimate both areas—are they really equal regardless of where you put the apex?

Proof Challenge

Available Justifications

1.

Given: Triangles ABC and DBC on the same base BC, between parallels AD and BC

Drag justification
2.

Through B, draw BE parallel to CA; through C, draw CF parallel to BD (using Prop 31)

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3.

EBCA and DBCF are parallelograms on the same base BC between the same parallels

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4.

Triangle ABC is half of parallelogram EBCA (diagonal BC bisects it)

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5.

Triangle DBC is half of parallelogram DBCF (diagonal BC bisects it)

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6.

Halves of equals are equal, so triangle ABC = triangle DBC in area

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 37.

  • ✓Lesson Plan (prop-37-lesson-plan.pdf)
  • ✓Student Worksheet (prop-37-worksheet.pdf)
  • ✓Answer Key (prop-37-answer-key.pdf)
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Why It Matters

This is the triangle version of Proposition 35. It establishes that a triangle's area depends only on its base and height (the distance between the parallels), not on the position of the apex. This is the geometric foundation of the formula 'area of a triangle = half base times height,' and it is used immediately in the Pythagorean theorem proof (Proposition 47).

Modern connection: The formula 'area = 1/2 base times height' is one of the first formulas students learn. Proposition 37 is its rigorous geometric justification. In calculus, this same idea — that area depends on base and perpendicular distance — generalizes to integration, where the area under a curve is computed as a limit of thin triangular and rectangular slices.

Historical note: Euclid's proof technique of 'doubling' triangles into parallelograms and then using Proposition 35 is a hallmark of Greek geometric reasoning. Rather than computing numerically, Euclid compares areas by embedding shapes inside larger, more tractable shapes. This 'completion to a parallelogram' technique recurs throughout Books I-IV.

Discussion Questions

  • The proof constructs parallelograms around each triangle and then takes halves. Why did Euclid not prove the triangle result directly, without going through parallelograms?
  • If you slide vertex A along the upper parallel line, the triangle ABC changes shape but its area stays constant. How does this connect to the concept of 'shearing' a shape?
  • This proposition says equal base and equal height imply equal area. The converse (Proposition 39) says equal area and same base imply equal height. Why is the converse harder to prove?
Euclid's Original Proof
Triangles which are on the same base and in the same parallels are equal to one another.

Let ABC, DBC be triangles on the same base BC and in the same parallels AD, BC;

I say that the triangle ABC is equal to the triangle DBC.

Let AD be produced in both directions to E, F;

through B let BE be drawn parallel to CA, [I.31]

and through C let CF be drawn parallel to BD. [I.31]

Then each of the figures EBCA, DBCF is a parallelogram; and they are equal,

for they are on the same base BC and in the same parallels BC, EF. [I.35]

Moreover the triangle ABC is half of the parallelogram EBCA; for the diameter AB bisects it. [I.34]

And the triangle DBC is half of the parallelogram DBCF; for the diameter DC bisects it. [I.34]

[But the halves of equal things are equal to one another.]

Therefore the triangle ABC is equal to the triangle DBC.

Therefore etc.

Q.E.D.

What's Next

Proposition 37 gives us the 'same base' case for triangles. Proposition 38 extends this to triangles with equal-length bases that don't need to coincide, mirroring exactly how Proposition 36 extended Proposition 35 for parallelograms.