20 Plane Rectilineal Geometry
Figure 1.8
D
B
Figure 1.10
sides. Hence Euclid needs 1,8 to infer the equality of the ap
propriate angles from the equality of the corresponding sides
of the two triangles. The procedure for constructing the triangle
.(4GF with sides equal to CE, ED, CD is given in proposition 22.
Euclid imagines an infinitely long straight line (fig. 1.8) on
which he marks off HA, AG, GI equal, respectively, to CD,
CE, ED. He then draws circles with centers A and G, radii AH
and GI, and argues that their intersection point F determines
with AG the desired triangle. But here the question of the exis
tence of F arises. It is intuitively clear that there would be no
such F if GI {DE) were not shorter than HG [CE CD), or
^//(CZ)) were not shorter than T/(C£' + ED),oy AG {CE)yNtrt
not shorter than HA -f GI {CD -f- DE). Hence, for 22 Euclid
needs 1,20 as a precondition {diorismos in the second sense of
this word).
I shall not here describe the sequence of propositions
leading to 1,20, but simply indicate the deductive relationships
diagramatically (fig. 1.9). The only logical play in this arrange
ment is provided by 14 and 17. 14 is not used until 1,45, but
its position is explained by its being the converse of 13. 17 is not
used until 111,16, but it is related to 1,16 in exactly the way that
the second part of 1,32 is related to the first. In this respect
17 is very like 28, which is not used until IV,7 and is related
to 27 in the same way that the last two parts of 29 are related
to the first part. It is not unreasonable to suppose that both
17 and 28 were inserted into the main deductive development
of book I when the subsequent need for them was realized.
Proposition 13 says in effect that if a straight line AB set
up on the straight line DC is not perpendicular to DC, then it
makes with DC angles equal to two right angles (fig. 1.10).
In the kataskeue of this proposition Euclid erects a perpendicular
EB to DC at B, which by definition 10 makes two right angles
with DC. The construction of the perpendicular is given by
1,11. In establishing 1,11 Euclid uses the bisection of an angle
(1,9). 1,10, the bisection of a straight line, is used in 1,16 and
42; but 1,12, the dropping of a perpendicular to a straight
line, is not invoked until 11,12. The sequence of constructions
1,9-12 has no real logical or geometric necessity. The four prob
lems could have been solved in any order, since they all depend
upon the fact that for a straight line s from the vertex of an
isosceles triangle to its base the following three conditions are
equivalent: s bisects the vertex angle; s bisects the base; s is
perpendicular to the base. Presumably the close relation of all
four accounts for Euclid’s proving them together, but there
seems to be no way to account for the particular sequence he
adopts.
21 Book I of the Elements
Figure 1.11
Figure 1.12
Figure 1.13
For future reference I give an alternative proof of 9-12
showing their interrelationship and their independence of
1,8, a proposition which Euclid invokes in the proofs of 9, 11,
and 12. In 9 one starts from an angle BAC, in 10 from a straight
line DE, in 11 from a straight line EG and a point H on it, and
in 12 from a straight line EG and a point A not on it (fig. 1.11).
For 9, 10, and 12 it is easy to construct an isosceles (and, for
10, equilateral) triangle with vertex A and base DE, and to
construct the equilateral triangle DIE on the other side of DE
from A. By 1,5 angle ADE equals AED and IDE equals lED.
Hence, angle .4 /)/equals AEI. By 1,4 angle equals EAH
(so DAE has been bisected); by 1,4 again, DH equals HE
(so DE has been bisected), and angle AHD equals AHE (so AH
is perpendicular to DE). For 11 one makes DH equal to HE,
constructs the equilateral triangle ADE, and argues by 1,4 that
angle AHD equals AHE.
1,4 is one of three “congruence” theorems for triangles
in book I, the others being 8 and 26.®® Only in 4 does Euclid
state the equality of the two triangles involved. Heath (vol.
I, p. 262), following Proclus (269.27-270.4), attributes this
difference to the fact that in 8 and 26 the equality of the triangles
is an easy inference from 1,4. However, the proof of 1,34 shows
that Euclid is conscious of the difference and does not take the
inference for granted.®® In 34 Euclid wishes to prove that if
AB, CD and AD, BC are two pairs of parallel straight lines,
then they are equal, and so are the angles DAB, BCD, and so
are the triangles ABD, DCB (fig. 1.12). Euclid infers the equal
ity of angle ABD to CDB and of CBD to ADB (and hence of
ABC to CD A) using 1,29. Since the side BD is common to the
two triangles, Euclid is able to infer the equality of the angles
DAB, BCD and of the pairs of parallel lines from 1,26. If
Euclid took 1,26 to include a tacit assertion of the equality of
the triangles, he would be finished. In fact, he argues for their
equality in terms of 1,4.
In general, Euclid’s treatment of the three congruence
theorems is quite perplexing. In 1,4 Euclid is given the equality
of AB, DE and of AC, DE and of angles BAC, EDF (fig. 1.13).
He proceeds by what is usually called the method of super
position {epharmodzein), a method depending upon intuitive
ideas about the coincidability of equal straight lines and angles
and upon the possibility of motion without deformation. Euclid
places the triangle ABC on DEE, making AB coincide with DE.
He then argues that AC and ZIF will coincide. From the coinci
dence of B, E and C, F Euclid concludes that BC and EF will
also coincide. This complete coincidence guarantees the equal
ity of these two sides, of the remaining pairs of angles, and of the
two triangles.