Proposition 45 of 48 • Construction

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

Given any rectilinear figure (polygon) and any angle, construct a single parallelogram that has exactly the same area as that polygon and has the given angle as one of its angles. This extends Proposition 42 (which handled triangles) to arbitrary polygons by triangulating the figure and applying areas one triangle at a time.

Before You Read

Can any polygon—a quadrilateral, a pentagon, an irregular hexagon—be converted into a single parallelogram of equal area with a chosen angle, using only compass and straightedge? The answer involves cutting the polygon into triangles and processing them one by one, but there is a real question: do the pieces snap together into one clean parallelogram, or do you get an awkward patchwork?

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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 rectilinear figure ABCD (any polygon) and a rectilinear angle E. We must construct a single parallelogram equal in area to the entire figure ABCD, with angle E as one of its angles.

Approach: Euclid triangulates the polygon by drawing diagonals. For the first triangle, he applies Proposition 42 to construct a parallelogram equal to it with the given angle. For each subsequent triangle, he uses Proposition 44 to apply a parallelogram equal to that triangle onto the end of the previous parallelogram, extending it. The key insight is that these parallelograms can be placed end-to-end on a single straight line because the angles match and the tops are all parallel (Proposition 29), so they combine into one large parallelogram.

Conclusion: Divide ABCD into triangles by drawing diagonal DB: triangle ABD and triangle DBC. Construct parallelogram FGHK equal to triangle ABD in angle E (Proposition 42). On line GH, apply parallelogram GHML equal to triangle DBC in angle E (Proposition 44). Since angle FKH and angle GHM each equal E, they are equal (CN1); adding angle KHG to both, angle FKH + angle KHG = angle KHG + angle GHM (CN2). But angle FKH + angle KHG = two right angles (parallels FK, GH with transversal KH — Proposition 29), so angle KHG + angle GHM = two right angles, and KH is in a straight line with HM (Proposition 14). By the same argument, FG is in a straight line with GL. Finally, FK is equal and parallel to HG, and HG to ML, so FK is equal and parallel to ML (Propositions 34 and 30); the straight lines KM and FL join them, so FKML is a parallelogram (Proposition 33) with angle FKM = E. Its area = FGHK + GHML = triangle ABD + triangle DBC = figure ABCD. ✓

Key Moves

Given: rectilinear figure ABCD and angle E.
Triangulate the figure: draw diagonal BD to get triangles ABD and BCD.
Construct parallelogram FGHK equal to triangle ABD in angle E (Proposition 42).
On straight line GH, apply parallelogram GHLM equal to triangle BCD in angle E (Proposition 44).
KH is in a straight line with HM: angle KHG + angle GHM = two right angles (Propositions 29 and 14).
Similarly FG is in a straight line with GL (Propositions 29 and 14).
FK is equal and parallel to HG, and HG to ML, so FK is parallel to ML (Propositions 34, 30); KM and FL join them, so FKML is a parallelogram (Proposition 33) with angle FKM = angle E.
Area of FKML = FGHK + GHLM = triangle ABD + triangle BCD = figure ABCD ✓

Try It Yourself

Draw an irregular quadrilateral and slice it into two triangles with a diagonal. Apply Proposition 42 to the first triangle to get a parallelogram with a right angle, then apply Proposition 44 to place the second triangle's area onto the end of that parallelogram. Verify that the two pieces share a straight boundary and form a single parallelogram.

Proof Challenge

Available Justifications

Given: Rectilinear figure ABCD and angle E. Decompose ABCD into triangles by drawing diagonal DB.

Drag justification

Construct parallelogram FGHK equal to triangle ABD in angle E.

Drag justification

Apply to line GH a parallelogram GHML equal to triangle DBC in angle E.

Drag justification

Angle FKH = angle GHM, since each equals the given angle E.

Drag justification

Add angle KHG to both: angle FKH + angle KHG = angle KHG + angle GHM.

Drag justification

Angle FKH + angle KHG = two right angles, since FK is parallel to GH with transversal KH.

Drag justification

So angle KHG + angle GHM = two right angles — therefore KH is in a straight line with HM. By the same argument at G, FG is in a straight line with GL.

Drag justification

FK is equal and parallel to HG, and HG to ML; so FK is equal and parallel to ML. The straight lines KM and FL join them at their endpoints — therefore FKML is a parallelogram.

Drag justification

Parallelogram FKML = parallelogram FGHK + parallelogram GHML = triangle ABD + triangle DBC = figure ABCD — and angle FKM = angle FKH = E (since KH lies along KM), so FKML is the required parallelogram in angle E.

Drag justification

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Curriculum Materials

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The lesson plan, student worksheet, and answer key for Proposition 45 come with the curriculum bundles.

✓Included in Advanced (Propositions 27–48)
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Why It Matters

This proposition completes the 'application of areas' program: any polygon whatsoever can now be converted into a single parallelogram of equal area with any prescribed angle. This is the most general area-transformation result in Book I and serves as the foundation for all subsequent area constructions in the Elements.

Going deeper

Modern connection: In modern terms, Proposition 45 shows that every polygon has a well-defined area that can be represented as a single parallelogram (or equivalently, as base times height). This is the geometric precursor to the idea that area is a single real number — a concept the Greeks achieved constructively without any number system for irrational quantities.

Historical note: The ability to reduce any polygon to a parallelogram of equal area was considered one of the great achievements of pre-Euclidean Greek geometry, attributed to the Pythagoreans by Proclus. It is the geometric version of a 'normal form' — reducing a complicated object to a standard shape for easier comparison.

Discussion Questions

The proof works by processing one triangle at a time. Could you process the triangles in any order, or does the order matter for the construction to work?
Proposition 45 handles any polygon. But the polygon must first be split into triangles. Is it always possible to triangulate a polygon using diagonals? Are there polygons where this is tricky?
This proposition, combined with Proposition 44, means you can build a parallelogram of prescribed area on any given base. How is this like solving the equation base times height = area?

Euclid's Original Proof

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

Let ABCD be the given rectilineal figure and E the given rectilineal angle; thus it is required to construct, in the given angle E, a parallelogram equal to the rectilineal figure ABCD.

Let DB be joined, and let the parallelogram FH be constructed equal to the triangle ABD, in the angle HKF which is equal to E; [I.42]

let the parallelogram GM equal to the triangle DBC be applied to the straight line GH, in the angle GHM which is equal to E. [I.44]

Then, since the angle E is equal to each of the angles HKF, GHM, the angle HKF is also equal to the angle GHM.

Let the angle KHG be added to each;

therefore the angles FKH, KHG are equal to the angles KHG, GHM.

But the angles FKH, KHG are equal to two right angles; [I.29]

therefore the angles KHG, GHM are also equal to two right angles.

Thus, with a straight line GH, and at the point H on it, two straight lines KH, HM not lying on the same side make the adjacent angles equal to two right angles;

therefore KH is in a straight line with HM. [I.14]

And, since the straight line HG falls upon the parallels KM, FG, the alternate angles MHG, HGF are equal to one another. [I.29]

Let the angle HGL be added to each;

therefore the angles MHG, HGL are equal to the angles HGF, HGL.

But the angles MHG, HGL are equal to two right angles; [I.29]

therefore the angles HGF, HGL are also equal to two right angles;

therefore FG is in a straight line with GL. [I.14]

And, since FK is equal and parallel to HG, [I.34]

and HG to ML also,

KF is also equal and parallel to ML; [C.N. 1; I.30]

and the straight lines KM, FL join them (at their extremities); therefore KM, FL are also equal and parallel. [I.33]

Therefore KFLM is a parallelogram.

And, since the triangle ABD is equal to the parallelogram FH, and DBC to the parallelogram GM,

the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM.

Therefore the parallelogram KFLM has been constructed equal to the given rectilineal figure ABCD, in the angle FKM which is equal to the given angle E.

(Being) what it was required to do.

What's Next

The application of areas is now complete. Proposition 46 steps back to prepare one final tool: how to construct a square on any given segment. That construction is the last ingredient Euclid needs before the climax of all of Book I—the Pythagorean theorem in Proposition 47.