HI 



MATHEMATICS. PLANE GEOMETRY. [BOOK H. PBOP. n. MT. 



CEB, E B C are each half a right angle, . . 

 A E B u a right angle. And because E B C U half 

 a right angle, DBG also is half a right angle, for 



u I. they are vertically opposite :t but B D G is a 

 right angle, being D CE, .'. the ivniniiiiiii; angle D(J B 

 u half a right angle, and is .-. -DBG, .'. BD- 



i. DO.* 



Again, because E G F U half a right angle, and that F 

 is a right angle, being the opposite angle E C D,t . . 



T si i. the remaining angle F E Q i half a riht 



I. aagle, .-. FEG -EOF, .-. GF - FE.* 

 And beeaoteE C- C A, . . E C 1 



- C A, . .E O + C A 1 - twice 

 CA 1 : but EA 1 - EC 1 + 



+ 471. CA,t .'. EA- 

 twioe AC 1 . Again, because 

 GF-FE. .-. GF'-FE 1 ,.-. 

 G F 1 + F E 1 - twice F E 1 ; but 



.4:1 EG'=GF 1 +FE 1 



* ML .-.EG 1 - twice FE 1 , and FE = CD,t .'. 

 EG 1 - twice C D 1 . But it was demonstrated that E A 1 



- twice C A 1 , /. AE 1 + EG 3 = twice A C 1 -)- twice CD-'; 

 .471 but AG 1 = AE' + EG 1 ,* /. AG 1 - twice 

 + 47L AC 1 -)- twice CD 1 ; but AD 1 + DG 1 =AG 1 ,t 



.-. A D 1 + D G 1 = twice AC 1 -)- twice C D s ; but D G = 

 DB, .-.AD 1 -r-DB 1 = etci A C" + hoic* C D 1 ; .-.ifa 

 straight line, &c. 



NOTE. For other demonstrations of the last two propositions, m the 

 rks at the end of this book. 



PROPOSITION XI PROBLEM. 



To divide a given straight line (A B) into two parts, so that 

 (tie rtetangle contained by the whole, and one of the park, 

 ihall be equal to the square of the other part. 



11*31. From A draw A C perpendicular and equal 



' I , 1 ' to A B ;* bisect A C in E,t and draw B E. 



Prolong C A to F, making E F = E B,* and 



from A B cut off A H = A F :t A B shall be divided in 



H, so that A B B H = A H 1 . 



MI Complete the square B C, as also the square 

 F H ;* and prolong G H to K. Then _ 



because A C is bisected in E and pro- 

 longed to F, the rectangle C F'F A + 

 ten. AE'-EF 1 ;-!- but EF = 



coiut. EB,* .-. CF-FA+AE* 



** EB; but BA + AE J = EB, be- *T~ 



cause E A B is a right angle, . . CF'FA 



+ A E 1 = B A 1 + A E s : take away 



A E 1 , which is common to both, . . 



CFFA-BA 1 . ButFK-CFFA, 



for A F = FG ; and AD = BA 1 , .-. 



FK=AD. Take away AK, common v * " 



to both, .-. FH -HD: but HD = AB'BH, for AB 



-BD, and F His AH 1 , .'. ABBH = A H J ; .-. AB 



isdividedinU, so that ABBH = AH 1 . Which was to 



be done. 



PROPOSITION XII. THEOREM. 

 In an obtuse-angled triangle (AB C), if a perpendicular 



(A D) be drawn from either of the acute angles to the 



<'fl>otite side prolonged, the square of the side (A B) sub- 

 tending the obtuse angle is greater 



than the squares of the sides (B C, A 



C A) containing the obt<ue angle, by 



turiot the rectangle contained by the 



side upon which, when prolonged, 



the perpendicular falls, awl the line, 



without the triangle, between the 



perpendicular and the obtuse angle ; 



that U, A B 1 - B C 1 + C A' + twice B C-C D. 



Because B D U divided into two parts in C, B D 1 



MIL BC' + CD' + twice BC-CD.* To each of 

 these add D A 1 .'. BD" + D A - BC + CD + D A 1 

 -f twice BC-CD; but AB 1 - B D' + DA 1 , andCA 1 - 



4471. C !) + ]> A*, t /.Ah 1 '- B O+ C A 1 + twice 

 BC-C D ; that is. the square of AB is greater than the 

 squares of BC, CA by twice B C C D ; .-. in an obtuse- 

 atualtd triangle, ic. Q. E. D. 



PROPOSITION XIII. THEOREM. 



In etwry triangle (A B C), Me square of a side (A C) sub- 

 tending an acute angle, is lest thin the squares of the 

 sides (C B, B A) containing that angle, by twice the rect- 

 angle (CB-BD) contained by 

 either of these sides, and the 

 straight line intercepted between 

 the perpendicular (A D) and the 

 acute angle (R) : that is, AC' 

 -C B 1 + B A" twice Cli It 1). 

 First, let A D f all within the tri- 

 angle ; then, because C B is 

 divided into two parts in D, C B 3 + B D J twice 



7H. CBBD + DC 1 .* To each of these add 

 AD 1 , ..09+BIP + AD*- twice CB-BD+ DC* 

 + A D* : but B A" - B D* + A I) 3 , and A C 1 - D C 1 



+ 471. +AD,t .'. CB + B A 1 - ACP + twice 

 CB-BD. From each of these take twice C B -B D, . . 

 AC 1 - Cn*+ B A 1 - twice CB'HD. 

 Secondly, let A D fall without the 

 triangle ; then, because D is a right 



16 1. angle, A C B is greater than 

 + 11 II. a right angle,* . . A B 1 = 



AC + CB* + twice BCCDt To 

 each of these add B C 2 , .-. AB" + 

 B C 1 - A C" + twice B C* + twice 

 BC-C D. Butbecause B D isdivided 

 into two parts in C, DBBC = BC'CD -f B C J ,* 



s ii. and the doubles of these being equal, . . A B ! 

 + B C 1 = A C 1 + twice D B B C. From each of these 

 take twice DB-BC, .'. AC 2 = A li' 4- BC 2 twice 

 D B-B C ; . ' . in every triangle, <tc. Q. E. D. 



PROPOSITION XIV. PROBLEM. 



To describe a square that shall be equal to a given rectilineal 

 figure (A). 



451. Describe the rectangle BD A.* Then, 



if the sides of it, B E, E D, are equal, it is a square, and 

 what was required is now done ; but if they are not equal, 

 prolong one of them, B E to F, and make E F = E D. 



+ lo i. Bisect B F in G,t and with centre G and 

 radius G B, or G F, describe the semicircle B H F, and 

 prolong DE to H. The 

 square described upon EH 

 shall be equal to the given 

 figure A. Draw G H. 

 Then because BF is di- 

 vided into two equal parts 

 in G, and into two unequal 

 parts in E, B E-E F+EG* 



5 II. = G F 2 ;* but 

 GF = GH, .-.BE-EF-r- 

 E G 2 = G H" ; but H E a 



+ 471. +EG 2 =GH2t BE-EF + EG, - H E 

 -f- EG*. Take away EG S , which is common to both, 

 .-. B E E F = E H 2 ; but B E'E F - B D, because E F 

 = ED, .-.BD = EH 2 ; but BD = A, .-. E H 2 = A ; 

 . . a square described tijmn E H u = to the given rectilineal 

 figure. Which was to be done. 



REMARKS ON BOOK II. 



All the propositions in this book, except the last 

 three, concern the divisions of straight lines into parts, 

 and the equality of rectangles having certain of these 

 parts for their sides. The first proposition is little more 

 than an axiom, as its object is to show that any rect- 

 angle B H is equal to the sum of the partial rectangles 

 1! K, D L, ic., into which it is divided: but, as already 

 observed (p. 545), Euclid does not regard as an axiom 

 any proposition that admits of demonstration. 



Proposition II. might, without impropriety, hare 

 been appended as a corollary to Proposition I., as 

 already stated in the text ; since, if the line A, in Pro- 

 jxwition I., be considered to be equal to the line BO, 

 this proposition will virtually include the next following. 



Proposition V. authorises us to afHrru, as an inference, 



