Proposition 38 of 48 • Theorem

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

If two triangles have equal-length bases and lie between the same pair of parallel lines, they have equal area — even if the bases are in different positions along the parallel and the triangles look quite different.

Before You Read

Two triangles with the same height but their bases at opposite ends of a field—does location matter, or is equal base-length enough to guarantee equal area? Euclid says no, location doesn't matter, but the proof requires a detour through a cleverly placed intermediate parallelogram.

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What Euclid Is Doing

Setup: Triangles ABC and DEF are on equal bases BC and EF, and between the same parallel lines. The bases need not coincide — they merely have equal length and lie on the same parallel. We must prove the triangles have equal area.

Approach: The strategy mirrors how Proposition 36 extended Proposition 35: construct parallelograms around each triangle, then use Proposition 36 (equal bases, same parallels) to show the parallelograms are equal, and Proposition 34 to show each triangle is half its parallelogram. Halves of equals are equal.

Conclusion: Through B, draw BG parallel to CA, and through E, draw EH parallel to DF (Proposition 31). Then GBCA and DEFH are parallelograms on bases BC and EF respectively, between the same parallels. Since BC = EF (given), by Proposition 36, parallelogram GBCA = parallelogram DEFH. The diagonal AC bisects GBCA (Proposition 34), so triangle ABC = half of GBCA. The diagonal DF bisects DEFH (Proposition 34), so triangle DEF = half of DEFH. Halves of equal things are equal (Common Notion 3). Therefore triangle ABC = triangle DEF. ✓

Key Moves

Given: triangles ABC and DEF on equal bases BC = EF, between the same parallels.
Through B, draw BG parallel to CA; GBCA is a parallelogram (Proposition 31).
Through E, draw EH parallel to DF; DEFH is a parallelogram (Proposition 31).
Parallelograms GBCA and DEFH have equal bases BC = EF and lie between the same parallels.
By Proposition 36, parallelogram GBCA = parallelogram DEFH.
Triangle ABC = half of GBCA, and triangle DEF = half of DEFH (Proposition 34).
Halves of equals are equal: triangle ABC = triangle DEF (Common Notion 3) ✓

Try It Yourself

Draw two separated triangles with equal-length bases on the same horizontal line and their apex on another horizontal line above—make one tall and skinny, the other with its apex far to one side. Measure or compute both areas. Then sketch the two parallelograms Euclid would construct around each triangle to complete the proof.

Proof Challenge

Available Justifications

Given: Triangles ABC and DEF on equal bases BC and EF, between parallels AD and BF

Drag justification

Through B, draw BG parallel to CA; through F, draw FH parallel to DE (using Prop 31)

Drag justification

GBCA and DEFH are parallelograms on equal bases BC and EF between the same parallels, so GBCA = DEFH in area

Drag justification

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

Drag justification

Triangle DEF is half of parallelogram DEFH (diagonal EF bisects it)

Drag justification

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

Drag justification

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 38.

✓Lesson Plan (prop-38-lesson-plan.pdf)
✓Student Worksheet (prop-38-worksheet.pdf)
✓Answer Key (prop-38-answer-key.pdf)

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Why It Matters

This completes the quartet of area propositions (35-38) by handling the most general case for triangles: equal bases, same height. Combined with Proposition 37, it establishes the full geometric basis for the formula 'triangle area = half base times height.' It is also needed for Proposition 40 (the converse).

Modern connection: In computational geometry and computer graphics, triangle area calculations are ubiquitous. Every polygon is triangulated, and areas are computed by summing triangle areas. The principle that only base and height matter — not the triangle's shape or position — is what makes these calculations efficient and position-independent.

Historical note: Propositions 35-38 form a tightly organized group: parallelogram same base (35), parallelogram equal bases (36), triangle same base (37), triangle equal bases (38). This systematic progression from specific to general, and from parallelograms to triangles, showcases Euclid's careful logical architecture.

Discussion Questions

How does this proposition relate to Proposition 37? Is Proposition 37 a special case of Proposition 38, or does each carry independent weight?
The proof depends on Proposition 36 (parallelograms on equal bases). Could you prove Proposition 38 directly from Proposition 37 without going through parallelograms?
Two triangles can have equal area without having equal bases or equal heights. This proposition gives a sufficient condition, not a necessary one. What other conditions guarantee equal area?

Euclid's Original Proof

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

Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD;

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

For let AD be produced in both directions to G, H;

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

and through F let FH be drawn parallel to DE. [I.31]

Then each of the figures GBCA, DEFH is a parallelogram;

and GBCA is equal to DEFH;

for they are on equal bases BC, EF and in the same parallels BF, GH. [I.36]

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

And the triangle FED is half of the parallelogram DEFH; for the diameter DF bisects it. [I.34]

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

Therefore the triangle ABC is equal to the triangle DEF.

Therefore etc.

Q.E.D.

What's Next

Props 37 and 38 prove that equal base and height imply equal area. The natural reversal—if two triangles have equal area and the same base, must their apexes be at the same height?—turns out to require proof by contradiction. That's the beautiful argument of Proposition 39.