---
book: 1
number: 44
id: "I.44"
kind: "construction"
uses: ["[[book-1/proposition-42]]", "[[book-1/proposition-31]]", "[[book-1/proposition-29]]", "[[book-1/proposition-43]]", "[[book-1/proposition-15]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.44"
license: "CC-BY-SA-4.0"
---

# I.44

To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.

![I.44](figures/I-44.svg)

## Proof

Let *AB* be the given straight line, *C* the given triangle and *D* the given rectilineal angle; thus it is required to apply to the given straight line *AB*, in an angle equal to the angle *D*, a parallelogram equal to the given triangle *C*.

Let the parallelogram *BEFG* be constructed equal to the triangle *C*, in the angle *EBG* which is equal to *D* [[book-1/proposition-42|I. 42]]; let it be placed so that *BE* is in a straight line with *AB*; let *FG* be drawn through to *H*, and let *AH* be drawn through *A* parallel to either *BG* or *EF*. [[book-1/proposition-31|I. 31]]

Let *HB* be joined.

Then, since the straight line *HF* falls upon the parallels *AH*, *EF*, the angles *AHF*, *HFE* are equal to two right angles. [[book-1/proposition-29|I. 29]] Therefore the angles *BHG*, *GFE* are less than two right angles; and straight lines produced indefinitely from angles less than two right angles meet; [[book-1/postulates#Postulate 5|Post. 5]] therefore *HB*, *FE*, when produced, will meet.

Let them be produced and meet at *K*; through the point *K* let *KL* be drawn parallel to either *EA* or *FH*, [[book-1/proposition-31|I. 31]] and let *HA*, *GB* be produced to the points *L*, *M*.

Then *HLKF* is a parallelogram, *HK* is its diameter, and *AG*, *ME* are parallelograms. and *LB*, *BF* the so-called complements, about *HK*; therefore *LB* is equal to *BF*. [[book-1/proposition-43|I. 43]]

But *BF* is equal to the triangle *C*; therefore *LB* is also equal to *C*. [[book-1/common-notions#Common Notion 1|C.N. 1]]

And, since the angle *GBE* is equal to the angle *ABM*, [[book-1/proposition-15|I. 15]] while the angle *GBE* is equal to *D*, the angle *ABM* is also equal to the angle *D*.

Therefore the parallelogram *LB* equal to the given triangle *C* has been applied to the given straight line *AB*, in the angle *ABM* which is equal to *D*.

Q. E. F.