Proposition 42 of 48 • Construction

To construct a parallelogram equal to a given triangle in a given rectilinear angle.

Given any triangle and any angle, build a parallelogram that has exactly the same area as the triangle and has the given angle as one of its angles. This is the first 'area conversion' construction — transforming a triangle into a parallelogram of equal area with a prescribed shape.

Before You Read

You have a triangle and an angle. Can you reshape that triangle into a parallelogram—same area, but with your chosen angle at one corner—using only compass and straightedge? It seems like converting between shapes while freezing the area should be tricky; the surprising answer is that it's entirely possible.

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What Euclid Is Doing

Setup: We are given a triangle ABC and a rectilinear angle D. We must construct a parallelogram that has the same area as triangle ABC and has angle D as one of its angles.

Approach: Euclid's key insight is to bisect the base BC at its midpoint E. This halves the base but keeps the height. He then constructs a parallelogram on the half-base EC with the required angle, and between the same parallels. Since the parallelogram on the half-base between the same parallels is double the triangle on the half-base (Proposition 41), and the original triangle on the full base equals two such triangles (Proposition 38), the areas match.

Conclusion: Bisect BC at E (Proposition 10). At point E on line EC, construct angle CEF equal to angle D (Proposition 23). Through A, draw AG parallel to EC (Proposition 31). Through C, draw CG parallel to EF (Proposition 31). Then FECG is a parallelogram with angle FEC = angle D. Now triangle AEC and parallelogram FECG share the same base EC and lie between the same parallels EC and AG. By Proposition 41, parallelogram FECG = 2 times triangle AEC. But triangle ABC has base BC = 2 times EC (since E is the midpoint), and triangles ABC and AEC share vertex A while base BC = 2 times base EC. Actually, since E is the midpoint of BC, triangle ABE = triangle AEC (Proposition 38, equal bases BE = EC, same height from A). So triangle ABC = 2 times triangle AEC. Therefore parallelogram FECG = 2 times triangle AEC = triangle ABC. The parallelogram equals the triangle. ✓

Key Moves

Given: triangle ABC and angle D.
Bisect BC at E (Proposition 10), so BE = EC.
At E, construct angle CEF = angle D (Proposition 23).
Through A, draw AG parallel to EC; through C, draw CG parallel to EF (Proposition 31).
FECG is a parallelogram with the required angle.
Parallelogram FECG and triangle AEC share base EC and lie between the same parallels.
By Proposition 41, parallelogram FECG = 2 times triangle AEC.
Since E is the midpoint of BC, triangle ABC = 2 times triangle AEC (Proposition 38).
Therefore parallelogram FECG = triangle ABC ✓

Try It Yourself

Draw any triangle and choose a target angle (try 45° or 60°). Now bisect the triangle's base, construct the angle at the midpoint, and complete the parallelogram by drawing the necessary parallels. Check that the result really has the same area as your triangle by placing one over the other.

Proof Challenge

Available Justifications

Given: Triangle ABC and angle D. Bisect BC at E.

Drag justification

At point E on line EC, construct angle CEF equal to angle D.

Drag justification

Through A, draw AG parallel to EC; through C, draw CG parallel to EF.

Drag justification

FECG is a parallelogram. Triangle AEB = triangle AEC (on equal bases BE = EC, same vertex A) — but we need: triangle AEB = triangle AEC since BE = EC and they share the same height.

Drag justification

So triangle ABC = 2 × triangle AEC. Parallelogram FECG = 2 × triangle AEC (same base EC, between same parallels). Therefore parallelogram FECG = triangle ABC.

Drag justification

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 42.

✓Lesson Plan (prop-42-lesson-plan.pdf)
✓Student Worksheet (prop-42-worksheet.pdf)
✓Answer Key (prop-42-answer-key.pdf)

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Why It Matters

This is the first area transformation construction in the Elements. It shows that any triangle can be converted into a parallelogram of equal area with any prescribed angle. This is a major step toward the 'application of areas' technique, which culminates in Propositions 44-45 and is one of the most powerful methods in Greek geometry.

Modern connection: In modern mathematics, this construction demonstrates that any triangle can be 'reshaped' into a parallelogram without changing its area — an early form of area-preserving transformation. In linear algebra, this relates to the idea that a linear map can transform any triangle into a parallelogram while preserving the absolute value of the determinant (the area scaling factor).

Historical note: Plutarch reportedly said that Pythagoras sacrificed an ox in celebration when he discovered the result behind this proposition — the ability to construct a parallelogram equal to a given triangle. While likely apocryphal, the story reflects how remarkable the ancient Greeks found this area transformation capability.

Discussion Questions

The construction bisects the base of the triangle. Why is halving the base the right strategy? What would go wrong if you tried to build the parallelogram on the full base?
The given angle D can be any angle at all. How does the shape of the resulting parallelogram change as angle D varies from very acute to very obtuse, while the area stays fixed?
This proposition converts a triangle to a parallelogram. Combined with Proposition 45, you can convert any polygon to a parallelogram. Why did the Greeks care so much about converting between shapes of equal area?

Euclid's Original Proof

To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.

Let ABC be the given triangle, and D the given rectilineal angle; thus it is required to construct in the rectilineal angle D a parallelogram equal to the triangle ABC.

Let BC be bisected at E, and let AE be joined;

on the straight line EC, and at the point E on it, let the angle CEF be constructed equal to the angle D; [I.23]

through A let AG be drawn parallel to EC, [I.31]

and through C let CG be drawn parallel to EF. [I.31]

Then FECG is a parallelogram.

And, since BE is equal to EC, the triangle ABE is also equal to the triangle AEC, for they are on equal bases BE, EC and in the same parallels BC, AG; [I.38]

therefore the triangle ABC is double of the triangle AEC.

But the parallelogram FECG is also double of the triangle AEC, for it has the same base with it and is in the same parallels with it; [I.41]

therefore the parallelogram FECG is equal to the triangle ABC.

And it has the angle CEF equal to the given angle D.

Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which is equal to the angle D.

(Being) what it was required to do.

What's Next

Proposition 42 converts a triangle into a parallelogram with any angle, but the resulting parallelogram sits wherever it lands—you have no control over which base it stands on. Proposition 44 solves that: it applies the same area to a specific prescribed base, which requires a clever use of the complement theorem coming up in Proposition 43.