Greeks and‘Origins’ 37
ADBCEFGFig. 2The diagram for Euclid proposition I.35.an example of the approach, is a single short proposition. The aim is a simple one: to show that two
parallelograms with the same base and the same height have equal area. In Euclid’s language, they
are just called ‘equal’.
PropositionI.35.Parallelogramswhichareonthesamebaseandinthesameparallelsequaloneanother.
Let ABCD and EBCF be parallelograms on the same base BC and in the same parallels AF and BC.
I say that ABCD equals the parallelogram EBCF. [Here is Euclid’s diagram (Fig. 2). The proof then
continues.]
Since ABCD is a parallelogram, therefore AD equals BC. (Proposition I.34.)
For the same reason EF equals BC, so that AD also equals EF. And DE is common, therefore the
whole AE equals the whole DF. (Common Notions 1, Common Notions 2.)
But AB also equals DC. Therefore, the two sides EA and AB equal the two sides FD and DC, respect-
ively, and the angle FDC equals the angle EAB, the exterior equals the interior. Therefore, the base
EB equals the base FC, and the triangle EAB equals the triangle FDC. (Propositions I.34, I.29, I.4.)
[So at this point, we have proved the triangles EAB and FDC are what is usually called ‘congruent’—
all sides and angles are the same. This implies that they have equal areas, of course.]
Subtract DGE from each. Then the trapezium ABGD which remains equals the trapezium EGCF.
(Common Notions 3.)
Add the triangle GBC to each. Then the whole parallelogram ABCD equals the whole
parallelogram EBCF. (Common Notions 2.)
Therefore, parallelograms which are on the same base and in the same parallels equal one
another. Q.E.D.
This result standsinstead ofthe numerical formula which we, like the ancient Egyptians and
Babylonians, would use to relate the area of a parallelogram (and, more commonly, a triangle), to
the length of the base and to the height.
This use of the term ‘equal’, to mean having the same area, is not in Euclid’s list of definitions;
you are supposed to know what it means. Here we see it as retrospectively defined by ‘adding’
and ‘subtracting’ the triangles DGE and GBC, starting from two triangles which themselves are
equal in all respects—all sides and all angles correspond. So, ‘equal’ means having equal area; but
paradoxically, ‘area’ is undefined. In a sense the process is a sleight of hand.
Notice also that the point G is not defined, except by the diagram; and also, that the proof as
stated only works if Gisas shown, that is, if CD crosses BE between the parallels. You can adjust
the proof for the case where it does not. Is this a ‘mistake’? Such minor oversights do occur, and