Euclid's WorkshopBook I
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Proposition 44 of 48 • Construction

To a given straight line in a given rectilinear angle, to apply a parallelogram equal to a given triangle.

Given a specific line segment, an angle, and a triangle, construct a parallelogram that has the given line segment as its base, the given angle as one of its angles, and exactly the same area as the given triangle. This is called 'application of areas' — you are applying (fitting) an area onto a prescribed base.

Before You Read

You are given a specific line segment, an angle, and a triangle. The challenge: construct a parallelogram that uses your line segment as its base, has your chosen angle at one corner, and has exactly the same area as the triangle. Can you always do this, even if the triangle is enormous and the base segment is tiny?

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C D A B E F G H K L M = area of C

What Euclid Is Doing

Setup: We are given a straight line AB, a rectilinear angle D, and a triangle C. We must construct a parallelogram that has AB as one side, angle D as one of its angles, and area equal to triangle C.

Approach: Euclid first uses Proposition 42 to construct any parallelogram BEFG equal to triangle C with angle D — but this parallelogram will generally not be on the line AB. The trick is to then extend and combine this parallelogram with line AB using the complement theorem (Proposition 43). By placing AB and the constructed parallelogram as complements of a larger parallelogram about its diagonal, the equal-complement property guarantees that the parallelogram on AB has the same area.

Conclusion: First, construct parallelogram BEFG equal to triangle C in angle D (Proposition 42), where BF is placed along AB extended. Place it so that BE is in line with AB. Now extend FE to meet AB produced, and complete a larger parallelogram containing both the line AB and BEFG. Draw the diagonal of this larger parallelogram. Through the intersection point, the lines parallel to the sides create complements: the parallelogram on AB and the parallelogram BEFG are complements about the diameter. By Proposition 43, these complements are equal. Since BEFG equals triangle C, the parallelogram on AB also equals triangle C. The angle at A is equal to angle D because the parallel lines preserve the angle (Proposition 29). ✓

Key Moves

  1. Given: line AB, angle D, triangle C.
  2. Construct parallelogram BEFG equal to triangle C in angle EBG = angle D (Proposition 42).
  3. Place BEFG so that BE lies along the extension of AB.
  4. Extend the figure to form a larger parallelogram with a diagonal passing through B.
  5. Through B, the larger parallelogram is divided into four parts: two 'about the diameter' and two complements (Proposition 31).
  6. The parallelogram on AB and BEFG are complements of the parallelograms about the diameter.
  7. By Proposition 43, the complements are equal: parallelogram on AB = BEFG.
  8. Since BEFG = triangle C, the parallelogram on AB = triangle C, with the required angle (Proposition 29) ✓

Try It Yourself

Draw a short base segment and a large triangle. Use Proposition 42 to first build a helper parallelogram equal to the triangle with the right angle, then try to 'slide' that area onto your prescribed base using the complement relationship. Watch how the diagonal of the extended figure enforces the area transfer.

Proof Challenge

Available Justifications

1.

Given: Line segment AB, triangle C, and angle D. Construct parallelogram BEFG equal to triangle C in angle D, with BE along AB extended.

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2.

Extend FG to H such that HB is parallel to FE. Join HB.

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3.

Through A, draw AH parallel to BG, and extend GB to meet it at K. HBAK is formed.

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4.

HBGK is a parallelogram with diagonal HB. Complements ABML and BEFG are about this diagonal. By Prop 43, complement ABML = complement BEFG.

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5.

Therefore parallelogram ABML = triangle C (since BEFG = triangle C), and angle ABM = angle D as required.

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 44.

  • ✓Lesson Plan (prop-44-lesson-plan.pdf)
  • ✓Student Worksheet (prop-44-worksheet.pdf)
  • ✓Answer Key (prop-44-answer-key.pdf)
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Why It Matters

This is the culmination of the 'application of areas' — one of the most celebrated techniques in Greek geometry. Given any line segment and any triangular area, you can build a parallelogram on that exact segment with that exact area. This means you can redistribute area to fit any prescribed base. Proposition 45 extends this to arbitrary polygonal areas, and the technique is the geometric ancestor of solving equations.

Modern connection: Application of areas is the geometric equivalent of solving a linear equation. If you think of the base as a given value 'b' and the area as a given value 'A,' then finding the parallelogram is like solving for the height h in the equation b times h = A. The Greeks had no algebra, but this construction achieves the same result purely geometrically. In modern algebraic geometry, this idea evolved into the theory of divisors and linear equivalence on algebraic curves.

Historical note: According to Proclus, the technique of 'application of areas' (parabolē) was invented by the Pythagoreans. It is one of three related operations: application (fitting area to a base exactly), application with excess (hyperbole), and application with deficiency (ellipse). These three Greek words — parabola, hyperbola, ellipse — were later adopted by Apollonius as names for the conic sections, which is how they got their modern names.

Discussion Questions

  • This construction combines Propositions 42 (build any equal parallelogram) and 43 (complements are equal) in a clever way. Why is it not enough to just use Proposition 42 alone?
  • The application of areas is essentially 'solving for height given base and area.' How does this compare to the algebraic approach of dividing area by base? What are the advantages and disadvantages of each?
  • Proclus tells us that the words parabola, hyperbola, and ellipse come from the area application technique. How does 'applying area to a line' relate to the shapes of conic sections?
Euclid's Original Proof
To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.

Let AB be the given straight line, C the given triangle and D the given rectilineal angle; thus it is required to apply to the given straight line AB, in an angle equal to the angle D, a parallelogram equal to the given triangle C.

Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D; [I.42]

let it be placed so that BE is in a straight line with AB;

let FG be drawn through to H, and let AH be drawn through A parallel to either BG or EF, [I.31]

and let HB be joined.

Then, since the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE are equal to two right angles. [I.29]

Therefore the angles BHG, GFE are less than two right angles;

and straight lines produced indefinitely from angles less than two right angles meet; [Post. 5]

therefore HB, FE, when produced, will meet. Let them meet at K.

Through the point K let KL be drawn parallel to EA or FH, [I.31]

and let HA, GB be produced to the points L, M.

Then HLKF is a parallelogram, HK is its diameter, and AG, ME are parallelograms, and LB, BF the so-called complements, about HK;

therefore LB is equal to BF. [I.43]

But BF is equal to the triangle C;

therefore LB is also equal to C. [C.N. 1]

And, since the angle GBE is equal to the angle ABM, [I.15]

while the angle GBE is equal to D,

the angle ABM is also equal to D.

Therefore the parallelogram LB equal to the given triangle C has been applied to the given straight line AB, in the angle ABM which is equal to D.

(Being) what it was required to do.

What's Next

Proposition 44 handles triangles; Proposition 45 breaks any polygon into triangles and applies Proposition 44 one piece at a time, snapping the pieces together into a single parallelogram. This completes the entire 'application of areas' program—any polygon, any angle, any base.