The Elements

...and to each of the angles at it, every side is equal to each, and the diameter cuts the areas in half. Let there be a parallelogram ΑΒΓΔ, and its diameter be ΑΓ; I say that the opposite sides and angles of the parallelogram ΑΒΓΔ are equal to each other. For since ΑΒ is parallel to ΓΔ and the straight line ΑΓ has fallen upon them, the alternate angles ΒΑΓ and ΑΓΔ are equal to each other. Again, since ΑΔ is parallel to ΒΓ and the straight line ΑΓ has fallen upon them, the alternate angles ΔΑΓ and ΑΓΒ are equal to each other. There are therefore two triangles, ΑΒΓ and ΑΓΔ, having two angles equal to two angles, each to each... which was to be demonstrated.

Parallelograms that are on the same base and between the same parallels are equal to each other.

Geometric diagram for Euclid's Proposition I.35, showing two parallelograms (ABCD and EBCF) sharing the same base line BC and situated between the same parallel horizontal lines. Vertices are labeled with the Greek letters Α, Β, Γ, Δ, Ε, Ζ.

In every parallelogram, the complements of the parallelograms around the diameter are equal to each other.

Geometric diagram for Euclid's Proposition I.43, depicting a large parallelogram divided by a main diagonal and lines parallel to the sides. It highlights the "complements" (parallelograms not on the diagonal) which the proposition proves to be equal. Labeled with letters Α, Β, Γ, Δ and a large Ω at the upper right corner.

Every full [set] of parallelograms that are on the same base and between the same parallels are equal to each other. Let the parallelograms ΑΒΓΔ and ΕΒΓΖ be on the same base ΒΓ and between the same parallels ΑΖ and ΒΓ. I say that ΑΒΓΔ is equal to ΕΒΓΖ. For since ΑΒΓΔ is a parallelogram, ΑΔ is equal to ΒΓ. For the same reasons, ΕΖ is also equal to ΒΓ. So ΑΔ is also equal to ΕΖ. And ΔΕ is common. Therefore, the whole ΑΕ is equal to the whole ΔΖ. And ΑΒ is equal to ΔΓ. Therefore, the two [sides] ΕΑ and ΑΒ are equal to the two ΖΔ and ΔΓ, each to each. And the angle ΕΑΒ is equal to the angle ΖΔΓ, the exterior to the interior. Therefore, the base ΕΒ is equal to the base ΖΓ, and triangle ΕΑΒ will be equal to triangle ΔΖΓ. Let the common triangle ΔΗΕ be subtracted; the remaining trapezium ΑΒΗΔ is therefore equal to the trapezium ΕΗΓΖ. Let the common triangle ΗΒΓ be added; the whole parallelogram ΑΒΓΔ is therefore equal to the whole parallelogram ΕΒΓΖ. Which was to be demonstrated.

Triangles that are on equal bases and between the same parallels are equal to each other. Let the triangles ΑΒΓ and ΔΕΖ be on equal bases ΒΓ and ΕΖ and between the same parallels ΑΔ and ΒΖ. I say that triangle ΑΒΓ is equal to triangle ΔΕΖ. For let ΑΔ be extended in both directions to Η and Θ, and through Β let ΒΗ be drawn parallel to ΑΓ, and through Ζ let ΖΘ be drawn parallel to ΔΕ. Therefore, each of ΗΒΓΑ and ΔΕΖΘ is a parallelogram. And they are equal; for they are on equal bases ΒΓ and ΕΖ and between the same parallels ΗΘ and ΒΖ. And triangle ΑΒΓ is half of the parallelogram ΗΒΓΑ; for the diameter ΑΒ cuts it in half. And triangle ΔΕΖ is half of the parallelogram ΔΕΖΘ; for the diameter ΔΖ cuts it in half. Halves of equal things are equal to each other. Therefore, triangle ΑΒΓ is equal to triangle ΔΕΖ. Which was to be demonstrated.

A series of three geometric triangles drawn side-by-side between two horizontal parallel lines, illustrating Euclid's Proposition I.38 (triangles on equal bases and in the same parallels are equal). Labels include Greek characters Α, Β, Γ, Δ, Ε, Ζ, Η, Θ.

A circular institutional stamp in the bottom right quadrant of the page, reading "BIBLIOTHECA REGIA MONACENSIS" surrounding a central emblem.

In every parallelogram, the complements of the parallelograms around the diameter are equal to each other. Let there be a parallelogram ΑΒΓΔ, and its diameter ΑΓ, and let ΕΘ and ΖΗ be the parallelograms around ΑΓ, and let ΒΚ and ΚΔ be the so-called complements; I say that the complement ΒΚ is equal to the complement ΚΔ. For since ΑΒΓΔ is a parallelogram, and its diameter is ΑΓ, triangle ΑΒΓ is equal to triangle ΑΔΓ...