26 



INTRODUCTION. 



that quantillos numerically incommenBurablc may be geometrically 

 determined. 



120. Theorem. Triangles and parallelograms of the 

 same height are proportional to their bases. 



Let AB be a common measure 

 of AC and AD, and let ABzzBE 

 =EF; join GB, GE, GF, tlicn 

 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 

 same multiple of AGB that AC is of AB, or AGD : AGB=:AD : AB, 

 and AGC ; AGBzzAC ; AB : hence, dividing the first equation by 

 the equal members of the second (18), AGD ; AGCzzAD : AC, and 

 2AGD : 2AGCzzAD : AC, therefore the parallelograms, which are 

 double the triangles, are also proportional. 



Scholium. The demonstration may easily be extended to in- 

 commensurable quantities. For if it be denied that AC : ADiz 

 AGC : AGD, let AC ; AD be the greater, and let the difference be 

 1 AGC n.AC AD «.AC— AD 



1 ., AC 



— , tlien 



» AD 



Letw.AD 



n AGD n.AT> n.AB n.AD 



be that multiple of AD wliich is less than n.AC, but greater than 



n.AC — AD, then a triangle on the base wi.AD will be equal to 



Jw.AGD, which will be less than n.AGC, the triangle on m.AC ; now 



n n.AGC w.AC— AD . 

 and 



multiplying tlie former equation by- , _ 



m m.AGD m.AD 



n.AGC.w.AD=:m.AGD. (w.AC— AD) ; but the first factors have been 

 shown to be respectiTely greater than the second, therefore tlieir pro- 

 ducts cannot be equal, and the supposition is impossible- 



121. Theorem. The homologous sides of equiangular 

 triangles are proportional. 



Let the homologous sides AB, BC, of tho 

 equiangular triangles ABD, BCE, be placed 

 contiguous to each other in the same line, 

 then AD 1 1 BE, and BD J | CE ; produce AD, 

 CE, till they meet in F, and join AE and BF. 

 Then the triangles FAE, EAC, are propor- 

 tional to their bases FE, EC, and the triangles AFB, BFC, to AB, 



