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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Basic Constructions (5)

Triangle Fundamentals (5)

Perpendiculars & Angles (5)

Exterior Angles & Inequalities (5)

Interior Triangles & Angle Copying (5)

Parallel Lines (5)

Parallel Constructions & Parallelograms (5)

Area Theorems (5)

Area Applications (4)

Grand Finale (4)

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 complete each triangle into a parallelogram, then compare the parallelograms. Produce AD in both directions; through B draw BE parallel to CA, and through C draw CF parallel to BD (Proposition 31), forming parallelograms EBCA and DBCF. Both share base BC and lie between the same parallels, so they are equal in area (Proposition 35). Each triangle is half its parallelogram — the diagonals AB and DC bisect them (Proposition 34) — so the triangles are equal.

Conclusion: Produce AD in both directions (Postulate 2). Through B, draw BE parallel to CA, meeting it at E; through C, draw CF parallel to BD, meeting it at F (Proposition 31). Now EBCA is a parallelogram (BE parallel to CA, EA parallel to BC), and DBCF is a parallelogram (BD parallel to CF, DF parallel to BC). Both are on the same base BC and between the same parallels, so parallelogram EBCA = parallelogram DBCF (Proposition 35). The diagonal AB bisects parallelogram EBCA (Proposition 34), so triangle ABC = half of EBCA. The diagonal DC bisects parallelogram DBCF (Proposition 34), so triangle DBC = half of DBCF. Halves of equal things are equal (the traditional companion of Common Notion 3). Therefore triangle ABC = triangle DBC. ✓

Key Moves

Given: triangles ABC and DBC on the same base BC, between the same parallels.
Produce AD in both directions (Postulate 2). Through B, draw BE parallel to CA, meeting it at E; EBCA is a parallelogram (Proposition 31).
Through C, draw CF parallel to BD, meeting the produced line at F; DBCF is a parallelogram (Proposition 31).
Parallelograms EBCA and DBCF share base BC and lie between the same parallels.
By Proposition 35, parallelogram EBCA = parallelogram DBCF.
Diagonal AB bisects EBCA: triangle ABC = half of EBCA (Proposition 34).
Diagonal DC bisects DBCF: triangle DBC = half of DBCF (Proposition 34).
Halves of equals are equal (traditional companion of Common Notion 3): triangle ABC = triangle DBC ✓

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

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

Drag justification

Produce AD in both directions. Through B, draw BE parallel to CA, meeting it at E; through C, draw CF parallel to BD, meeting it at F

Drag justification

EBCA and DBCF are parallelograms on the same base BC between the same parallels, so EBCA = DBCF in area

Drag justification

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

Drag justification

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

Drag justification

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

Drag justification

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Curriculum Materials

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

Going deeper

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.