14 



OF SPACE. 



~o, therefore 3 is the common ttjeasure, for it measures 9, 

 and 9+3 or 12, and ia+9 or 21, and 2X21 + 12 or it. 



Scholium. Hence it is obvious, that there can be no 

 greater common measure of the two quantities than the 

 quantity thus found ; for it should measure tl>e difference 

 of the two quantities, and all the successive remainders 

 down to the last, therefore it cannot be greater than this 

 last. It must also be remarked, that in some cases no ac- 

 curate common measure can be f*nnd, but the error, or 

 the last remainder, in this process, itiay always be less than 

 any quantity that can be assigned, since the process may 

 be continued without limit. That there are incommensu- 

 rable quantities, may be thus shown : every number Is 

 either a prime number, that is, a number not capable of 

 being composed by multiplication of other numbers, or \t 

 is composed by the multiplication of factors, which are 

 primes. Let the number a be composed of the prime 

 numbers led, or az^bcd, then aa':^Lcd,tcd'^zlh.cc,dd and 

 each prime factor of aa occurs twice ; so that every square 

 number must be composed of factors in pairs ; and a square 

 number multiplied by a number which is not composed of 

 factors in pairs cannot be a square number : for instance, 

 2aa or 3aa cannot be a square number, since the factors 

 of 2 are only 1.2, and of 3, 1.3, and not in pairs : there- 

 fore the square rooot of 2 or 3 cannot be expressed by any 

 fractiofi, for the square of its numerator would be twice or 

 th-ice the square of its denominator. But the ratio of the 

 hypotenuse of a triangle to its side may be that of »/ 2 or 

 4/3 to 1 ; so that quantities numerically incommensurable 

 may be geometrically determined. 



IGO. Theorem. Triangles and paiallelo- 

 grams of the same height are proportional to 

 their bases. 



Let AB be a 

 common measure 

 of AC and AD, 

 and let AB=:BE 



A B E r C 'J) =Efi J°'" GB, 



'GE, G¥, then the triangles AGB, BGE, EGF, are equal, 

 and the triangle AGD is the same multiple of AGB that 

 AD is of AB ; and AGC-is the sanre multiple of AGB that 

 AC is of AB, or AGD : AGB=AD : AB, and AGC : AGB 

 rzAC ; AB ; hence, dividing the first equation by the equal 

 terms of the second (is), AGD : AGCnAD : AC, and 

 aAGD : 2AGC=:AD : AC, therefore the parallelograms 

 ■which are double the triangles, are also proportional. 



ScHOLi u M. The demonstration may easily be extended 

 •to incommensurable quantities. For if it be denied that 

 AC : AD=:AGC : AGD, let AC : AD be the greater, and 



■ , ,.^ , I , AC 1 AGC n.AC 



let the difference be — , then — : :r -— — — 



n AD n AGD n.AU 



AD 



' n.AW 



n.AC-AV> 



. Let m.AD be that multiple of 

 n.AD ' 



AD which is less than n.AC, but greater than ?i.AC 

 — AD, then a triangle on the base mAD will be equal 

 to m.AGD, which will be less than n.AGC, the tri- 

 angle on n.AC ; now multiplying the former equation by 



rt 71. AGC 7J.AC— AD , ,„„ ,„ .^„ 



— , r: , and 7i.AGC.7n..\D=m.AGD. 



m m.AGD m.AD 



(ji.AC — ^AD); btjt the first factors have been shown to be 

 respectively greater than the second, therefore their pro- 

 ducts cannot be equal, and the supposition is impossible. 



l<21. Theorem; The homologous sides 

 oFetjuianguIar triangles are proporliioual. 



Let the homologous sides AB,BC, p 

 of the equiangular triangles ABD, 

 BCE, be placed contiguous to each 

 other in the same line, then AD 1 1 

 BE,and BDl ICE; produce AD.CE, ABC 



till they meet in F, and join AB and BF. Then the tri- 

 angles FAE, EAC, are proportional to their bases FE, EC, 

 and the triangles AFB, BFC, to AB, BC (120). But FAE 

 =:AFB(ll7),andEAC=EBC4-EAB=EBC+EFB=BFC, 

 therefore FAE : EAC=AFB : BFC, and FE : EC=AB : BC; 

 but FEi^DB (114). In the same manner it may be shown 

 that the other homologous sides are proportional. 



ScHOLrUM. Hence equiangular triangles are also called 

 similar. 



122. Theorem. Equal and equiangular 

 parallelograms have their sides reciprocally 

 proportional. 



If AB=:BC then DB : BE=BF : j^ q. 



BG. ForDB:BF=.AB:GF(l2o;=: f 



BC : GF=BE : BG (120) ; or DB : D 

 BErzBF : BG. 



£/ 



E C 



123. Theorem. Equiangular parallelo- 

 grams, having their sides reciprocally pro- 

 portional, are equal. 



For they may be placed as in the last proposition, and 

 the demonstration will be exactly similar. 

 ■ Scholium. Hence is derived the common method of 

 finding the contents of rectangles ; let a and b be the sides 

 of a rectangle, then 1 : a: il: al, and the rectangle is equal 

 to that of which the sides are 1 and ab, or to ab square units. 

 Hence the rectangle contained by two lines is equivalent to 

 the product of their numeral representatives. 



