---
book: 1
number: 34
id: "I.34"
kind: "theorem"
uses: ["[[book-1/proposition-29]]", "[[book-1/proposition-26]]", "[[book-1/proposition-4]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.34"
license: "CC-BY-SA-4.0"
---

# I.34

In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.

![I.34](figures/I-34.svg)

## Proof

Let *ACDB* be a parallelogrammic area, and *BC* its diameter; I say that the opposite sides and angles of the parallelogram *ACDB* are equal to one another, and the diameter *BC* bisects it.

For, since *AB* is parallel to *CD*, and the straight line *BC* has fallen upon them, the alternate angles *ABC*, *BCD* are equal to one another. [[book-1/proposition-29|I. 29]]

Again, since *AC* is parallel to *BD*, and *BC*has fallen upon them, the alternate angles *ACB*, *CBD* are equal to one another. [[book-1/proposition-29|I. 29]]

Therefore *ABC*, *DCB* are two triangles having the two angles *ABC*, *BCA* equal to the two angles *DCB*, *CBD* respectively, and one side equal to one side, namely that adjoining the equal angles and common to both of them, *BC*; therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle; [[book-1/proposition-26|I. 26]] therefore the side *AB* is equal to *CD*, and *AC* to *BD*, and further the angle *BAC* is equal to the angle *CDB*.

And, since the angle *ABC* is equal to the angle *BCD*, and the angle *CBD* to the angle *ACB*, the whole angle *ABD* is equal to the whole angle *ACD*. [[book-1/common-notions#Common Notion 2|C.N. 2]] And the angle *BAC* was also proved equal to the angle *CDB*.

Therefore in parallelogrammic areas the opposite sides and angles are equal to one another.

I say, next, that the diameter also bisects the areas.

For, since *AB* is equal to *CD*, and *BC* is common, the two sides *AB*, *BC* are equal to the two sides *DC*, *CB* respectively; and the angle *ABC* is equal to the angle *BCD*; therefore the base *AC* is also equal to *DB*, and the triangle *ABC* is equal to the triangle *DCB*. [[book-1/proposition-4|I. 4]]

Therefore the diameter *BC* bisects the parallelogram *ACDB*.

Q. E. D.