KESPECTING THE THEORY OF PARALLELS. 447 



it, the straight line which bisects the sides of the one bisects also the sides of the 

 other. 



For both these lines pass through LM, the points of bisection of the saone 

 isosceles triangle. 



Cor. 3. Euclid's Prop. 39, " Equal triangles upon the same base, and upon 

 the same side of it, are between the same parallels, " is true. 



For the perpendiculars from the vertices of the equal triangles on GH are 

 both equal to KI, and therefore equal to one another. 



Hence the line joining their vertices is parallel to GH ; but GH is parallel to 

 BC ; therefore the line joining the vertices is parallel to the base. 



The last part of the demonstration holds good, from the circumstance that 

 the straight line GH, to which the two are both parallel, lies between them. 

 Euclid's Proposition 30 is not admissible, and has not been assumed. It will be 

 seen by Prop. IX. that Euclid's proposition 30 is not compatible with our axiom. 

 Hence also the converse of this corollary is not true. Instead of it, i. e., instead 

 of Euclid's Prop. 37, we have the following. 



Coi\ 4. Triangles upon the same base, and whose other sides are bisected by 

 the same straight line, are equal to one another. 



Let ABC, KBC, be any two triangles on the same base, and having their 

 other sides bisected by the same straight line GH, area ABC = KBC. 



For each triangle is equal to the figure GBCH. 



Cor. 5. Euclid's Prop. 40, "Equal triangles on the same side of bases which 

 are equal and in the same straight line are between the same parallels, " is true. 

 The demonstration differs little from that given above for Cor. 3. 



Cor. 6. Euclid's Prop. 38 requires a similar alteration to that made on Prop. 

 37 in Cor. 4. 



Cor. 7. If I be mil/ point in the line GH, the triangle KBC is still equal both 

 in area and in the sum of its angles to ABC. 



Cor. 8. Also KM = MC as before. 



Cor. 9. The results given in Cors. 7 and 8 are also true if L be taken, any 

 point in GH, and LK be made equal to BL, for then KI = BG as before. 



Cor. 10. Hence, upon BC there can be described a triangle equal in area and 

 in the sum of its angles to ABC, and having one side equal to a given straight 

 line greater than one of the sides AB, BC, of the triangle. For BL may be taken 

 equal to half that line. 



Prop. XXI. To find a pomt in the base of a triangle iwoduced, such that the tri- 

 angle on the part produced, of which the vertex is the same as that of the given 

 t7'iangle shall he equal in area to the given triangle ; when it is possible. 



Let ABC be the given triangle ; C the angle which is not greater than B, and 



y 



