KESPECTING THE THEORY OF PARALLELS. 



437 



Prop. V. If two triangles have two angles of the one equal to two angles of the other, 

 each to each, hut the side adjacent to the equal angles of the one greater than 

 the side adjacent to the equal angles of the other, then the other sides of the 

 triangle which has the greater adjacent side are respectively greater than the 

 other sides of the other triangle, hut they contain a less angle. 



Let^ ABC = DEF, and^ ACB = DFE, but BC:::-EF, then is B A::^ED, and AC:^DF. 

 Cut off from BC, BH = EF; and make the angle BHG=EFD or BCA; HG shall 

 cut BA in G ; for if it do not cut B A it must cut CA, and make with HC and CA 

 a triangle, having the exterior angle equal to the interior opposite angle, which is 

 impossible. Now (Euc. I. 26) BG=ED, but BA^-BG .-. BA^ED. Similarly AC 

 ^DF. Also the angle BAC is less than EDF. 



For BGH + BHG + AGH + CHG = four right 

 angles ; but GAC + ACH + AGH + CHG are less 

 than four right angles (Axiom Cor. 2) ; there- 

 fore BGH:^BAC, and BGH = EDF, therefore EDF; 



'BAC. 



Prop. VI. If from two points without a straight line, on the same side of it, the two 

 perpendicidars drawn to the line are equal, the st7^aight line which joins the 

 points is parallel to the given li^ie. 



Let AB be the given straight line, C, D the given points without it on the same 

 side, from which the two equal perpendiculars are drawn 

 to the line, viz., CE, DF; CD is parallel to EF. 



Join CF, ED meeting in G ; and through G draw GH 

 perpendicular to AB, and produce HG to meet CD in K. 



The triangles CEF, DFE, are equal in every respect, 

 therefore CF = DE, and^CFE = DEF: hence the triangles GHE, GHF are equal 

 (Euc. I. 26), therefore EG=FG, and ^EGH = FGH: hence, by subtraction, 

 CG=GD; and ^CGK = DGK, therefore (Euc. I. 4) ^CKG=DKG, and each 

 of them is a right angle; i.e., HK is at right angles to both the lines AB and CD; 

 wherefore CD is parallel to AB. 



Cor. 1. The same is true if CE, DF are not perpendicular but equally inclined 

 to AB ; for then perpendiculars from C and D to AB are equal. 



Cor. 2. HK bisects the sides CD, EF, and the area CEFD. 



Cor. 3. A perpendicular equidistant from two equal perpendiculars is a 

 common perpendicular to two parallels, found as above. 



VOL. XXIIL PART IIL 6 C 



