---
book: 6
number: 10
id: "VI.10"
kind: "construction"
uses: ["[[book-1/proposition-31]]", "[[book-1/proposition-34]]", "[[book-6/proposition-2]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.10"
license: "CC-BY-SA-4.0"
---

# VI.10

*To cut a given uncut straight line similarly to a given cut straight line*.

## Proof

Let *AB* be the given uncut straight line, and *AC* the straight line cut at the points *D*, *E*; and let them be so placed as to contain any angle; let *CB* be joined, and through *D*, *E* let *DF*, *EG* be drawn parallel to *BC*, and through *D* let *DHK* be drawn parallel to *AB*. [[book-1/proposition-31|I. 31]]

Therefore each of the figures *FH*, *HB* is a parallelogram; therefore *DH* is equal to *FG* and *HK* to *GB*. [[book-1/proposition-34|I. 34]]

Now, since the straight line *HE* has been drawn parallel to *KC*, one of the sides of the triangle *DKC*, therefore, proportionally, as *CE* is to *ED*, so is *KH* to *HD*. [[book-6/proposition-2|VI. 2]]

But *KH* is equal to *BG*, and *HD* to *GF*; therefore, as *CE* is to *ED*, so is *BG* to *GF*.

Again, since *FD* has been drawn parallel to *GE*, one of the sides of the triangle *AGE*, therefore, proportionally, as *ED* is to *DA*, so is *GF* to *FA*. [[book-6/proposition-2|VI. 2]]

But it was also proved that, as *CE* is to *ED*, so is *BG* to *GF*; therefore, as *CE* is to *ED*, so is *BG* to *GF*, and, as *ED* is to *DA*, so is *GF* to *FA*.

Therefore the given uncut straight line *AB* has been cut similarly to the given cut straight line *AC*. Q. E. F.