BOOK I. PROP. I IV.] 



MATHEMATICS. PLANE GEOMETRY. 



643 



And that circle may be described from any centre, at 

 any distance from that centre. 



AXIOMS. 



i. 



Things which are equal to the same thing are equal to 

 one another. 



ii. 

 If equals be added to equals, the wholes will be equal. 



ii r. 



If equals be taken from equals, the remainders will be 

 equal 



rv. 



If equals be added to unequals, the wholes will be 

 unequal. 



v. 



If equals be taken from unequals, the remainders will 

 be unequal. 



VI. 



Things which are double of the same are equal to one 

 another. 



VII. 



Things which are halves of the same are equal to one 

 another. 



vitl. 



Magnitudes whicn coincide, or which may be conceived 

 to coincide, with one another, are equal. 



The whole is greater than its part. 



Zi 

 Two straight lines cannot enclose a space. 



XI. 



Ail right angles are equal to one another. 



HARKS AlfD SIONS CSED FOR ABBREVIATION. 



= is the sign of equality, and signifies that the quan- 

 tities between which it is placed are equal. 



+ is the sign of addition, and signifies that the 

 quantities between which it ia placed are to be added. 



ia the sign of subtraction, and implies that the 

 quantity after it, is to be subtracted from the quantity 

 before it. 



.*. stands for the word therefore, or consequently. 



Besides these marks the following contractions are also 

 frequently used ; namely, ax. for axiom ; post, for pos- 

 tulate ; prob. for problem ; theo. for theorem ; prop, for 

 proposition ; const, for construction ; and hyp. for hypo- 

 thesis. Q. E. D. stands for " Quod erat demonstrandum;" 

 that is, which was to be demonstrated. 



PROPOSITION L PROBLEM. 



To describe an equilateral triangle (A B C) upon a given 

 finite straight line (A B). 



With centre A, and radius AB, describe the circle 



Post. 3. BCD;* and with centre B, and the sam e 

 radius, B A, describe the circle 



ACE. 



From C, the point in which 

 the circles cut each other, draw 



Po.t.1. CA, CB,* the 

 triangle ABC shall be equila- 

 teral. 



t Def. is. Because A is the 



AX. i. centre of the 



circle BCD, AC = AB,t and because B is the centre 

 of the circle ACE, BC = AB, /. AC = BC;* so that 

 AC, B C, A B, are equal to one another ; consequently 

 tii# triangle A B C is equilateral, and it is described upon 

 A B, which was to be done. 



c _. 



PROPOSITION II. PROBLEM. 



From a given point (A) to draw a straight line (A L) eq-Md 

 to a given straight line (B C). 



Post 1. From A to B draw AB,* and upon AB describe 



Pr. i. the equilateral triangle B D A.* 'With 

 centre B, and radius B C, describe the circle 



t Post. s. C G H,t and pro- 

 duce DB to meet it in G. Again: 

 with D as centre, and D G as 

 radius, describe the circle 

 G K L, and produce D A to 



Post. 2. meet it in L ; * 

 A L shall be equal to B C. 



Because B is the centre of 

 the circle C G H, B C = 



t Def. IS. BG.f 



Because D is the centre of 

 the circle G K L, D L = D G. 



But DA, DB, parts of D L, D G, are also 



Const. equal,* . . the remainders, A L, B G, are 

 Ax. 3. equal ;t so that A L, B C are each equal to 



B G, and are, consequently, equal to each other.* 



Ax. 1. Wherefore, from the given point, A, a 

 straight line, A L, has been drawn equal to B C. Wliich 

 was to be done. 



PROPOSITION III. PROBIEM. 



From the greater (A B) of two given straight lines (A B 

 and C), to cut off apart equal to the less (C). 



Pr. l. From A draw AD = C,* and with centre 

 A, and radius A 1), de- 

 scribe the circle D E F,t 



+ Post. s. cutting off 

 A E from A B : A E shall 

 be = C. 



Because A is the centre 

 of the circle DEF, AK= 



Der. u. AD.* But 

 t Const. C=AD,f.'. 



A K and C are each = 



AT. i. A 1), . -. A E=C ; * so that, from the greater 

 line, A B, a part A E, has been cut off, equal to the less, 

 C. If-VitcA was to be done. 



PROPOSITION IV. THEOREM, 



// two triangles (A B C, I) E F) have two sides (A B, A C) 

 of the one equal to two sides (D E, D F) of the other, 

 each to each ; that is, A B = D E, and A C = D F, and 

 have likewise the angles contained by those sides equal, 

 that it, the angle A = the angle D, then their bases or 

 third sides (B C, E F) shall be equal, and also their other 

 angles each to each, namely, those to which the equal 

 sides are opposite, that is, the angle B = the angle E, 

 and the angle C = the angle F ; the surfaces of the tri- 

 angles shall also be equal. 

 For, conceive the triangle A B C to be laid upon, or 



applied to, the triangle D E F so that the point A may 



be on D, and the line 



AB onDE; then the A. 



point B shall fall on 



E, because A B = 



Hyp. DE. And 

 A B thus coinciding 

 with D E, AC must 

 fall on DF, because the 

 angle A = the angle , 



tuyp. D;t also the 



Hyp. point C must fall on F, because A C = D F. * 

 But B was proved to coincide with E ; and since B coin- 

 cides with E, and C with F, the base B C must coincide 

 with the base E F, otherwise the two straight lines, B C, 

 E F, thus made to coincide at their extremities, would 



t AX. 10. inclose a space, which is impossible ; t ' the 

 base B C, coinciding with the base E F, is equal to 



AX. 8. E F. * Moreover, since the two lines, A B, 

 B C coincide with the two D E, E F, the angle B coin- 

 cides with the angle E ; and since the two lines AC, C B 

 coincifle with the two D F, F E, the angle C coincides 

 with the angle F ; .'. the angle B = the angle E, and the 



