---
book: 1
number: 43
id: "I.43"
kind: "theorem"
uses: ["[[book-1/proposition-34]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.43"
license: "CC-BY-SA-4.0"
---

# I.43

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

![I.43](figures/I-43.svg)

## Proof

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* its diameter, the triangle *ABC* is equal to the triangle *ACD*. [[book-1/proposition-34|I. 34]]

Again, since *EH* is a parallelogram, and *AK* is its diameter, the triangle *AEK* is equal to the triangle *AHK*. For the same reason the triangle *KFC* is also equal to *KGC*.

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*. [[book-1/common-notions#Common Notion 2|C.N. 2]]

And the whole triangle *ABC* is also equal to the whole *ADC*; therefore the complement *BK* which remains is equal to the complement *KD* which remains. [[book-1/common-notions#Common Notion 3|C.N. 3]]

Therefore etc.

Q. E. D.