---
book: 1
number: 42
id: "I.42"
kind: "construction"
uses: ["[[book-1/proposition-23]]", "[[book-1/proposition-31]]", "[[book-1/proposition-38]]", "[[book-1/proposition-41]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.42"
license: "CC-BY-SA-4.0"
---

# I.42

To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.

![I.42](figures/I-42.svg)

## Proof

Let *ABC* be the given triangle, and *D* the given rectilineal angle; thus it is required to construct in the rectilineal angle *D* a parallelogram equal to the triangle *ABC*.

Let *BC* be bisected at *E*, and let *AE* be joined; on the straight line *EC*, and at the point *E* on it, let the angle *CEF* be constructed equal to the angle *D*; [[book-1/proposition-23|I. 23]] through *A* let *AG* be drawn parallel to *EC*, and [[book-1/proposition-31|I. 31]] through *C* let *CG* be drawn parallel to *EF*.

Then *FECG* is a parallelogram.

And, since *BE* is equal to *EC*, the triangle *ABE* is also equal to the triangle *AEC*, for they are on equal bases *BE*, *EC* and in the same parallels *BC*, *AG*; [[book-1/proposition-38|I. 38]] therefore the triangle *ABC* is double of the triangle *AEC*.

But the parallelogram *FECG* is also double of the triangle *AEC*, for it has the same base with it and is in the same parallels with it; [[book-1/proposition-41|I. 41]] therefore the parallelogram *FECG* is equal to the triangle *ABC*.

And it has the angle *CEF* equal to the given angle *D*.

Therefore the parallelogram *FECG* has been constructed equal to the given triangle *ABC*, in the angle *CEF* which is equal to *D*. Q. E. F.