---
book: 6
number: 24
id: "VI.24"
kind: "theorem"
uses: ["[[book-6/proposition-2]]", "[[book-5/proposition-18]]", "[[book-5/proposition-16]]", "[[book-5/proposition-22]]", "[[book-6/proposition-21]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.24"
license: "CC-BY-SA-4.0"
---

# VI.24

*In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another*.

## Proof

Let *ABCD* be a parallelogram, and *AC* its diameter, and let *EG*, *HK* be parallelograms about *AC*; I say that each of the parallelograms *EG*, *HK* is similar both to the whole *ABCD* and to the other.

For, since *EF* has been drawn parallel to *BC*, one of the sides of the triangle *ABC*, proportionally, as *BE* is to *EA*, so is *CF* to *FA*. [[book-6/proposition-2|VI. 2]]

Again, since *FG* has been drawn parallel to *CD*, one of the sides of the triangle *ACD*, proportionally, as *CF* is to *FA*, so is *DG* to *GA*. [[book-6/proposition-2|VI. 2]]

But it was proved that, as *CF* is to *FA*, so also is *BE* to *EA*; therefore also, as *BE* is to *EA*, so is *DG* to *GA*, and therefore, *componendo*, as *BA* is to *AE*, so is *DA* to *AG*, [[book-5/proposition-18|V. 18]] and, alternately, as *BA* is to *AD*, so is *EA* to *AG*. [[book-5/proposition-16|V. 16]]

Therefore in the parallelograms *ABCD*, *EG*, the sides about the common angle *BAD* are proportional.

And, since *GF* is parallel to *DC*, the angle *AFG* is equal to the angle *DCA*; and the angle *DAC* is common to the two triangles *ADC*, *AGF*; therefore the triangle *ADC* is equiangular with the triangle *AGF*.

For the same reason the triangle *ACB* is also equiangular with the triangle *AFE*, and the whole parallelogram *ABCD* is equiangular with the parallelogram *EG*.

Therefore, proportionally, as *AD* is to *DC*, so is *AG* to *GF*, as *DC* is to *CA*, so is *GF* to *FA*, as *AC* is to *CB*, so is *AF* to *FE*, and further, as *CB* is to *BA*, so is *FE* to *EA*.

And, since it was proved that, as *DC* is to *CA*, so is *GF* to *FA*, and, as *AC* is to *CB*, so is *AF* to *FE*, therefore, ex aequali, as *DC* is to *CB*, so is *GF* to *FE*. [[book-5/proposition-22|V. 22]]

Therefore in the parallelograms *ABCD*, *EG* the sides about the equal angles are proportional; therefore the parallelogram *ABCD* is similar to the parallelogram *EG*. [[book-6/definitions#Definition 1|VI. Def. 1]]

For the same reason the parallelogram *ABCD* is also similar to the parallelogram *KH*; therefore each of the parallelograms *EG*, *HK* is similar to *ABCD*.

But figures similar to the same rectilineal figure are also similar to one another; [[book-6/proposition-21|VI. 21]] therefore the parallelogram *EG* is also similar to the parallelogram *HK*.

Therefore etc. Q. E. D.