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**Proposition I.36. ^{1}] Parallelograms which are on equal bases and in the same parallels equal one another.^{2}**

Let *ABCD* and *EFGH* be parallelograms which are on the equal bases *BC* and *FG* and in the same parallels *AH* and *BG*.

I say that the parallelogram *ABCD* equals *EFGH*.^{3}

Join *BE* and *CH*.^{4}

Since *BC* equals *FG* and *FG* equals *EH*, therefore *BC* equals *EH*.

But they are also parallel, and *EB* and *HC* join them. But straight lines joining equal and parallel straight lines in the same directions are equal and parallel, therefore *EBCH* is a parallelogram.^{5}

And it equals *ABCD*, for it has the same base *BC* with it and is in the same parallels *BC* and *AH* with it.^{6}

For the same reason also *EFGH* equals the same *EBCH*, so that the parallelogram *ABCD* also equals *EFGH*.

Therefore parallelograms which are on equal bases and in the same parallels equal one another. QED^{7}

**[Proposition I.38.] Triangles which are on equal bases and in the same parallels equal one another.**

Let *ABC* and *DEF* be triangles on equal bases *BC* and *EF* and in the same parallels *BF* and *AD*.

I say that the triangle *ABC* equals the triangle *DEF*.

Produce *AD* in both directions to *G* and *H*. Draw *BG* through *B* parallel to *CA*, and draw *FH* through *F* parallel to *DE*.

Then each of the figures *GBCA* and *DEFH* is a parallelogram, and *GBCA* equals *DEFH*, for they are on equal bases *BC* and *EF* and in the same parallels *BF* and *GH*.

Moreover the triangle *ABC* is half of the parallelogram *GBCA*, for the diameter *AB* bisects it. And the triangle *FED* is half of the parallelogram *DEFH*, for the diameter *DF* bisects it.

Therefore the triangle *ABC* equals the triangle *DEF*.

Therefore triangles which are on equal bases and in the same parallels equal one another. qed

**[Proposition I.42] To construct a parallelogram equal to a given triangle in a given rectilinear angle.^{8}**

Let *ABC* be the given triangle, and *D* the given rectilinear angle.

It is required to construct in the rectilineal angle *D* a parallelogram equal to the triangle *ABC*.

Bisect *BC* at *E*, and join *AE*. Construct the angle *CEF* on the straight line *EC* at the point *E* on it equal to the angle *D*.^{9} Draw *AG* through *A* parallel to *EC*, and draw *CG* through *C* parallel to *EF*.

Then *FECG* is a parallelogram.

Since *BE* equals *EC*, therefore the triangle *ABE* also equals the triangle *AEC*, for they are on equal bases *BE* and *EC* and in the same parallels *BC* and *AG*.^{10} Therefore the triangle *ABC* is double the triangle *AEC*.

But the parallelogram *FECG* is also double the triangle *AEC*, for it has the same base with it and is in the same parallels with it,^{11} therefore the parallelogram *FECG* equals 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 equals *D*. QEF^{12}

**Footnotes:**

^{1} The text is Sir Thomas Heath's well-known English translation. We adopt the common convention of referring to propositions from the *Elements* by book and number, whence I.36 refers to the 36th proposition of Book I.

^{2}Notice that it is the *areas* of the parallelograms that are being discussed here; Euclid makes no distinction between the area of the figure and the figure itself. Further, the statement of the proposition makes it clear that the area of a parallelogram depends only on the length of its base and its corresponding height (the distance between the parallels to which reference is made in the statement of the proposition). This is clear to the modern geometer when she writes that the formula for the area of a parallelogram with base *b* and height *h* is *A* = *bh*. But Euclid is not a modern geometer, so he sees no need to make reference to the (numerical) height of the parallelogram when the purely geometric form of the pair of parallels enclosing the figure is description enough for him. Euclid does not *compute* the area of the figure, he simply characterizes it in terms of its "elements."

^{3}It will be obvious from the style of the writing, here and throughout the *Elements*, that Euclid adopts a very formal style of writing, almost like poetic verse, to lay out geometric propositions. The initial statement of the proposition, given in italics in the text, is called the *protasis*. It identifies the premises or conditions and the corresponding conclusion of the proposition at hand, often (though not in this instance) in the form "If ... , then ... " The protasis is followed by a second version of the hypotheses of the proposition, the *ekthesis*, this time with letters referring to a specific, but entirely generic, illustration of the given data. The ekthesis is also accompanied by a diagram to which the letters refer. This is immediately followed by the *diorismos*, or restatement of the conclusion, often using the words "I say that ... " The remaining parts of the format for Euclidean propositions will be discussed in the next note.

^{4}The fourth part of the propostion is called the *kataskeu¯e* (it need not appear in all propositions). It calls for any additional lines or curves that should be added to the diagram to help effect the proof. It is followed by the *apodeixis*, the meat of the proof, which lists the chain of reasons leading from the given premises to the final conclusion.

^{5}Euclid calls here on the fact, which he has proved earlier (in Proposition I.33) that in a quadrilateral, if one pair of opposite sides is parallel and equal, then so must the other pair be.

^{6}Euclid also cites an earlier proposition here to justify this step of the argument. But it is easy enough to see the reason directly from the diagram: since triangles *ABE* and *BCH* have equal dimensions (see the figure), then they are clearly equal, and removing each of these triangles in turn from the trapezoid *ABCH* leaves the parallelograms *ABCD* and *EBCH*, which must therefore also be equal.

^{7}The final sentence of the proposition is called the *sumperasma*; it is simply a recapitulation of the original statement of the proposition, prefaced with a "therefore," and typically followed by the phrase "which was to be demonstrated" (in Latin, *quod erat demonstrandum*, often abbreviated as QED).

^{8}The form of the statement of this proposition is different from that of the two previous propositions we have displayed here. Whereas Propositions I.36 and I.38 are what are technically called *theorems*, or assertions of mathematical fact, Proposition I.42 is what the Greeks called a *problem*, in which the goal is to find some desired point, line, figure, or number. All of Euclid's propositions have one of these two forms.

^{9}Euclid, in I.23, showed how to construct a copy of a given angle along a given line at any particular point on it. It is a straightforward construction procedure.

^{10}You recognize this as precisely the content of Proposition I.38.

^{11}Euclid has established this - that the triangle whose base is the same as a given parallelogram and whose opposite vertex lies on the extensions of the opposite and parallel side of the parallelogram is equal to half the parallelogram - in Proposition I.41. In other words, if a triangle has base *b* and height *h*, and a parallelogram has the same dimensions, then the area of the triangle (*A* = [1/2]*bh*) is exactly half that of the parallelogram (*A* = *bh*).

^{12}Since this proposition is a problem, not a theorem, the sumperasma ends with the phrase "which was to be constructed," in Latin, "*quod erat factum*."

Daniel E. Otero (Xavier University), "The Quadrature of the Circle and Hippocrates' Lunes - The Quadrature of Polygons in Euclid's [i]Elements,[/i] Book I," *Convergence* (July 2010)