196 A History ofMathematics
while Proclus agreed that to prove the postulate one would need extra definitions and theorems, he
did not say that one would need an alternative postulate. One needed, essentially, to provide a way
of seeing why what did not appear self-evident could be made so. Such were the terms of reference,
again until the seventeenth century. The recurring problem is that what is self-evident to one writer
may be illegitimate to another; and that almost anyone who made a serious attempt agreed that the
problem was difficult (necessarily, or it would not have preoccupied so many eminent predecessors)
and accordingly a proof could not be simple.
In a detailed analysis reproduced as Fauvel and Gray 6.C.3, Youschkevitch describes the various
proofs from the Islamic period and their flaws. As an example of one which seems quite adequate,
we could consider ibn al-Haytham’s proof, reproduced as Fauvel and Gray 6.C.1. Khayyam criticises
it (loc.cit. 6.C.2) for using the idea of motion, but in fact this is not the major problem. Euclid used
ideas of motion implicitly in places, and Th ̄abit ibn Qurra in discussing the same question gave a
reasoned defence. Here is how ibn al-Haytham begins:Let us start with a premise for that, and that is: ‘When two straight lines are produced from the extremities of a finite
straight line, containing two right angles with the first line, then every perpendicular dropped from one of these two
lines on the other is equal to the first line, which contained two right angles with these two lines.’ [The meaning must
be (see Fig. 5) that the angles actually are two right angles, not as sometimes in Euclidean-language that they add to
two right angles—otherwise the statement is obviously untrue.] Thus, every perpendicular dropped from one of the
afore-mentioned lines on the other contains a right angle with the line from which it was dropped. An example of this
is as follows: there is extended from the two extremities of lineABtwo linesAG,BD, and the anglesGAB,DBAare each
right. Then pointGis assumed on lineAGand from it perpendicularGDis dropped on lineBD. I say, then, that lineGD
is equal to lineAB. The proof of that is that nothing else is possible. (Fauvel and Gray, p. 235)Notice first, that ibn al-Haytham states clearly what he is using as a replacement for the parallel
postulate. (It is again equivalent, since it essentially asserts the existence of rectangles.) Second, he
does not consider it self-evident, despite the bold ‘nothing else is possible’; because he goes on to
prove his statement from what hedoesconsider self-evident, or at least more basic geometry. The
details are quite complex; you can follow them through in the source and find, if you can, where
something ‘equivalent’ to postulate 5 is being used as an assumption.Exercise 1.(a) How does the ‘angles of a triangle’ theorem’ follow from the parallel postulate in either
of the forms cited above? (b) LetABbe a straight line; constructAC,BDperpendicular toABand on the
same side of it, so thatAC=BD. What would you need to show (1) that the angles at C and D are right
angles, (2) thatAB=CD?Exercise 2.Prove the ‘Claim’ above about the distance of the meeting-point.AGBD
Fig. 5Ibn al-Haytham’s method—basically to construct a rectangle. Angles A, B, and G are right angles; the statement is then that
AG must be equal to BD. (We shall find Lambert discussing quadrilaterals of this type much later.)