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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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: 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 the angle EBG = D (Proposition 42), placed so that BE is in a straight line with AB. Produce FG to H, draw AH through A parallel to BG (Proposition 31), and join HB. Since AH is parallel to EF with transversal HF, the angles AHF + HFE equal two right angles (Proposition 29); so the angles BHG + GFE are less than two right angles, and HB, FE produced must meet (Postulate 5) — say at K. Through K draw KL parallel to EA, and produce HA to L and GB to M (Proposition 31). Now HLKF is a parallelogram with diameter HK, B lies on HK, AG and ME are the parallelograms about the diameter, and ABML and BEFG are the complements — so ABML = BEFG (Proposition 43). Since BEFG equals triangle C, parallelogram ABML also equals triangle C (Common Notion 1). Finally, angle ABM = angle GBE (vertical angles, Proposition 15) = angle D. ✓

Key Moves

Given: line AB, angle D, triangle C.
Construct parallelogram BEFG equal to triangle C in angle EBG = angle D (Proposition 42), with BE in a straight line with AB.
Produce FG to H; through A draw AH parallel to BG (Proposition 31); join HB.
Angles AHF + HFE = two right angles (Proposition 29), so BHG + GFE < two right angles — HB and FE produced meet at K (Postulate 5).
Through K draw KL parallel to EA; produce HA to L and GB to M (Proposition 31).
HLKF is a parallelogram with diameter HK through B; AG and ME are about the diameter; ABML and BEFG are the complements.
By Proposition 43, complement ABML = complement BEFG = triangle C (Common Notion 1).
Angle ABM = angle GBE (vertical angles, Proposition 15) = angle D ✓

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

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

Drag justification

Extend FG through G to H, where H is the point at which the line drawn through A parallel to BG meets it. Join HB.

Drag justification

Since AH is parallel to EF with transversal HF, angle AHF + angle HFE = two right angles. So angle BHG + angle GFE < two right angles — therefore HB and FE, when produced, meet at a point K.

Drag justification

Through K, draw KL parallel to EA; produce HA to L and GB to M.

Drag justification

HLKF is a parallelogram with diameter HK, and B lies on HK. AG and ME are the parallelograms about HK; ABML and BEFG are the complements. By Prop 43, complement ABML = complement BEFG.

Drag justification

Therefore parallelogram ABML = triangle C (since BEFG = triangle C).

Drag justification

Angle ABM = angle GBE (vertical angles at B) = angle D. So ABML is the required parallelogram on AB in the given angle.

Drag justification

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Curriculum Materials

Get the Teaching Materials

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

Going deeper

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.