---
book: 1
number: 45
id: "I.45"
kind: "construction"
uses: ["[[book-1/proposition-42]]", "[[book-1/proposition-44]]", "[[book-1/proposition-29]]", "[[book-1/proposition-14]]", "[[book-1/proposition-34]]", "[[book-1/proposition-30]]", "[[book-1/proposition-33]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.45"
license: "CC-BY-SA-4.0"
---

# I.45

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

![I.45](figures/I-45.svg)

## Proof

Let *ABCD* be the given rectilineal figure and *E* the given rectilineal angle; thus it is required to construct, in the given angle *E*, a parallelogram equal to the rectilineal figure *ABCD*.

Let *DB* be joined, and let the parallelogram *FH* be constructed equal to the triangle *ABD*, in the angle *HKF* which is equal to *E*; [[book-1/proposition-42|I. 42]] let the parallelogram *GM* equal to the triangle *DBC* be applied to the straight line *GH*, in the angle *GHM* which is equal to *E*. [[book-1/proposition-44|I. 44]]

Then, since the angle *E* is equal to each of the angles *HKF*, *GHM*, the angle *HKF* is also equal to the angle *GHM*. [[book-1/common-notions#Common Notion 1|C.N. 1]]

Let the angle *KHG* be added to each; therefore the angles *FKH*, *KHG* are equal to the angles *KHG*, *GHM*.

But the angles *FKH*, *KHG* are equal to two right angles; [[book-1/proposition-29|I. 29]] therefore the angles *KHG*, *GHM* are also equal to two right angles.

Thus, with a straight line *GH*, and at the point *H* on it, two straight lines *KH*, *HM* not lying on the same side make the adjacent angles equal to two right angles; therefore *KH* is in a straight line with *HM*. [[book-1/proposition-14|I. 14]]

And, since the straight line *HG* falls upon the parallels *KM*, *FG*, the alternate angles *MHG*, *HGF* are equal to one another. [[book-1/proposition-29|I. 29]]

Let the angle *HGL* be added to each; therefore the angles *MHG*, *HGL* are equal to the angles *HGF*, *HGL*. [[book-1/common-notions#Common Notion 2|C.N. 2]]

But the angles *MHG*, *HGL* are equal to two right angles; [[book-1/proposition-29|I. 29]] therefore the angles *HGF*, *HGL* are also equal to two right angles. [[book-1/common-notions#Common Notion 1|C.N. 1]] Therefore *FG* is in a straight line with *GL*. [[book-1/proposition-14|I. 14]]

And, since *FK* is equal and parallel to *HG*, [[book-1/proposition-34|I. 34]] and *HG* to *ML* also, *KF* is also equal and parallel to *ML*; [[book-1/common-notions#Common Notion 1|C.N. 1]];[[book-1/proposition-30|I. 30]] and the straight lines *KM*, *FL* join them (at their extremities); therefore *KM*, *FL* are also equal and parallel. [[book-1/proposition-33|I. 33]] Therefore *KFLM* is a parallelogram.

And, since the triangle *ABD* is equal to the parallelogram *FH*, and *DBC* to *GM*, the whole rectilineal figure *ABCD* is equal to the whole parallelogram *KFLM*.

Therefore the parallelogram *KFLM* has been constructed equal to the given rectilineal figure *ABCD*, in the angle *FKM* which is equal to the given angle *E*.

Q. E. F.