A History of Mathematics- From Mesopotamia to Modernity

208 A History ofMathematics

BC

E

A

F

D

G

Fig. 11The picture for Euclid I.16.

B

C

D

E

A

F

Fig. 12A ‘large’ triangle on a sphere, and the way in which proposition I.16 fails.

Bisect AC at E. Join BE, and produce it in a straight line to F. Make EF equal to BE, join FC, and

draw AC through to G (see Fig. 11)

Since AE equals EC, and BE equals EF, therefore the two sides AE and EB equal the two sides CE

and EF, respectively, and the angle AEB equals the angle FEC, for they are vertical angles. Therefore,

the base AB equals the base FC, the triangle ABE equals the triangle CFE, and the remaining angles,

equal the remaining angles, respectively, namely those opposite the equal sides. Therefore, the

angle BAE equals the angle ECF.

But the angle ECD is greater than the angle ECF, therefore the angle ACD is greater than the

angle BAE. Similarly, if BC is bisected, then the angle BCG, that is, the angle ACD, can also be proved

to be greater than the angle ABC. Therefore in any triangle, if one of the sides is produced, then the

exterior angle is greater than either of the interior and opposite angles. Q.E.D.

This propositionfailsin the ‘geometry’ of shortest lines on a sphere. In fact, it is not

hard to construct triangles in which it is untrue; one such is shown in Fig. 12. Where does

the proof fail? The point F constructed above no longer falls ‘inside’ the angle ACD, as seems

obvious from the Euclidean picture; and so the angle ECD is no longer greater than the angle

ECF. Hence, although the previous steps in the proof work, the crucial use of ‘greater than’

does not.