Euclid's WorkshopBook I
← Back to Propositions
Proposition 43 of 48 • Theorem

In any parallelogram the complements of the parallelograms about the diameter equal one another.

Draw a diagonal of any parallelogram and then draw lines through any point on that diagonal parallel to the sides. This creates four smaller parallelograms. The two that the diagonal passes through are 'about the diameter.' The other two — the 'complements' — always have equal area, no matter where the point on the diagonal is chosen.

Before You Read

Pick any point on the diagonal of a parallelogram and draw two lines through it—one parallel to each pair of sides. You get four smaller parallelograms. The diagonal threads through two of them; the other two are the 'leftovers.' Are those two leftover pieces always equal in area, regardless of where you placed the point on the diagonal?

Browse All Foundations

All Foundations

Definitions (23)
Postulates (5)
Common Notions (5)
Browse All Propositions

All Propositions

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)
A B C D K E F G H = =

What Euclid Is Doing

Setup: Let ABCD be a parallelogram with diagonal AC. Let K be any point on AC. Through K, draw lines parallel to the sides AB and AD. This divides the parallelogram into four smaller parallelograms. The two that the diagonal AC passes through (the ones containing A and C as corners) are the 'parallelograms about the diameter.' The other two are the 'complements.' We must show the complements have equal area.

Approach: Euclid uses Proposition 34 (a diagonal bisects a parallelogram) applied three times: once to the whole parallelogram and once to each of the two smaller parallelograms about the diameter. By subtracting the 'about the diameter' triangles from the whole-diagonal triangles, the complements emerge as equal remainders.

Conclusion: The diagonal AC bisects the whole parallelogram ABCD: triangle ABC = triangle ACD (Proposition 34). Within the top-left smaller parallelogram about the diameter, diagonal AK bisects it: triangle AEK = triangle AFK (Proposition 34). Within the bottom-right smaller parallelogram about the diameter, diagonal KC bisects it: triangle KGC = triangle KHC (Proposition 34). Now, triangle ABC = triangle AEK + complement 1 + triangle KGC. And triangle ACD = triangle AFK + complement 2 + triangle KHC. Since triangle ABC = triangle ACD, and triangle AEK = triangle AFK, and triangle KGC = triangle KHC, subtracting the equal parts from equal wholes leaves equal remainders (Common Notion 3): complement 1 = complement 2. ✓

Key Moves

  1. Given: parallelogram ABCD with diagonal AC, point K on AC, lines through K parallel to the sides.
  2. Diagonal AC bisects ABCD: triangle ABC = triangle ACD (Proposition 34).
  3. The diagonal also passes through the two smaller parallelograms 'about the diameter.'
  4. AK bisects the first small parallelogram: triangle AEK = triangle AFK (Proposition 34).
  5. KC bisects the second small parallelogram: triangle KGC = triangle KHC (Proposition 34).
  6. Triangle ABC = triangle AEK + complement 1 + triangle KGC.
  7. Triangle ACD = triangle AFK + complement 2 + triangle KHC.
  8. Subtract equal triangles from equal wholes: complement 1 = complement 2 (Common Notions 1-3) ✓

Try It Yourself

Draw a parallelogram and its diagonal, then pick a point on the diagonal and draw the two parallel lines through it. Label the four pieces and estimate their areas. Now move your point closer to one end of the diagonal and repeat—does the equality of the two complements hold no matter where the point sits?

Proof Challenge

Available Justifications

1.

Given: Parallelogram ABCD with diagonal AC. Points E, F on the diagonal determine sub-parallelograms AEKH and KFCG about the diagonal.

Drag justification
2.

Triangle ACD = triangle ACB (diagonal AC bisects parallelogram ABCD)

Drag justification
3.

Triangle AEK = triangle AHK (diagonal AK bisects sub-parallelogram AEKH)

Drag justification
4.

Triangle KFC = triangle KGC (diagonal KC bisects sub-parallelogram KFCG)

Drag justification
5.

Triangle ACD − triangle AHK − triangle KGC = triangle ACB − triangle AEK − triangle KFC. Therefore complement HKCD = complement ABKH — i.e., the complements are equal.

Drag justification
0 of 5 steps completed

Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 43.

  • ✓Lesson Plan (prop-43-lesson-plan.pdf)
  • ✓Student Worksheet (prop-43-worksheet.pdf)
  • ✓Answer Key (prop-43-answer-key.pdf)
Browse Curriculum Bundles

Why It Matters

This elegant result is the key lemma for Proposition 44 (application of areas). The ability to slide area around within a parallelogram — from one complement to the other — while preserving equality is what makes it possible to construct a parallelogram on a given base equal to a given figure. Without Proposition 43, the powerful 'application of areas' technique would not work.

Modern connection: In matrix theory, the complements of a partitioned matrix have a related structure. In computational geometry, Proposition 43 underlies algorithms for area-preserving mesh transformations. The result also appears in the geometric proof that every parallelogram can be dissected into sub-parallelograms with predictable area relationships.

Historical note: This proposition is sometimes called the 'Gnomon theorem' because the L-shaped region formed by the two complements and one 'about the diameter' parallelogram is called a gnomon. Gnomons were important in Greek mathematics — they appear in Pythagorean number theory (as the odd numbers that build square numbers) and in the geometric algebra of Book II.

Discussion Questions

  • The complements are always equal regardless of where K sits on the diagonal. What happens to their shape and size as K moves from A toward C? When are the complements largest?
  • This proposition is purely about area — the complements are equal in area but generally different in shape. Could they ever be congruent (same shape and size)? Under what conditions?
  • Why does Euclid need this result? It seems abstract on its own. How does knowing that complements are equal help with constructing parallelograms of prescribed area (Proposition 44)?
Euclid's Original Proof
In any parallelogram the complements of the parallelograms about the diameter are equal to one another.

Let ABCD be a parallelogram, and AC its diameter; and about AC let EH, FG be parallelograms, and BK, KD the so-called complements;

I say that the complement BK is equal to the complement KD.

For, since ABCD is a parallelogram, and AC is its diameter, the triangle ABC is equal to the triangle ACD. [I.34]

Again, since EH is a parallelogram, and AK is its diameter, the triangle AEK is equal to the triangle AHK. [I.34]

For the same reason the triangle KFC is equal to the triangle KGC. [I.34]

Now, since the triangle AEK is equal to the triangle AHK, and KFC to KGC,

the triangle AEK together with KGC is equal to the triangle AHK together with KFC. [C.N. 2]

And the whole triangle ABC is also equal to the whole triangle ACD;

therefore the complement BK which remains is equal to the complement KD. [C.N. 3]

Therefore etc.

Q.E.D.

What's Next

This complement equality is not just a pretty symmetry—it is the hidden mechanism that makes Proposition 44 work. By recognizing the 'applied' parallelogram and a helper parallelogram as complements inside a larger figure, Euclid can guarantee their areas are equal and transfer any triangular area onto a prescribed base.