OF SPACE. 



II 



with ABC, For if it be denied, let G be the vertex of the 

 triangle so constructed ; join CG ; then since AC^AG, 

 ^ACG^AGC (87), and in the same manner /.BGCr: 

 bCG; but BGOACG, therefore BGOACG; and 

 ACG>BCG, therefore much more BGOBCG, to which 

 it was shown to be equal. And the same may be proved 

 in any other position 6f the point G ; therefore the triangle 

 equal to DEF, supposed to be described on AB, coincides 

 with ABC. 



90. PROBtEM. To bisect a given angle. 



A In the right lines forming the angle, take 



^ at pleasure ABi^AC; on BC describe an 

 ' equilateral triangle BCD, and AD will bisect 

 the angle BAC. For ABz:AC, BD=CD, 

 and the base AD is common, therefore the 

 triangle ABDrzACD (89), and z.BAD= 

 CAD. 



Dl. Problem. To bisect a given right 

 line, AB. 



Describe on it two equilateral triangles, 

 ABC, ABD, and CD, joining their ver- 

 tices, will bisect AB in E. For since 

 ACzzOB, AD=BD, and CD is common 

 to the triangles ACD, BCD, /.ACDz: 

 BCD (sa) ; but CE is common to the 

 triangles ACE and BCE, therefore AE= 

 EB (86). 



92. Problem. To erect a [x^rpendicular 

 to a given right Hne at a given point. 



On each side the point A, take at plea- 

 sure ABziAC, and on BC make an equi- 

 lateral triangle, BCD. Then AD shall 

 be perpendicular to BC. For the sides 



of BAD and CAD are respectively equal. 



Bag therefore the angle BADzzCAD (89), and 

 both are right angles (6!), and AD is perpendicular to 

 BC (»5,\ 



93. Problem. From a point, A, without 

 a right line^ BC, to let fall a perpendicular 

 on it. 



On the centre A, through any point 



D, beyond BC, describe a circle, which 



E \^must obviously cut BC; join AB and 



BC. 



C AC, and bisect the angle BAC by the 

 line AE ; AE will be perpendicular to 

 For ^BAE=CAE, AB:::AC, and AE is common 



to the triangles BAE, CAR; therefore z.AEBr:AEC (88\ 

 and both are right angles (64). 



94. Theorem. The angles which any 

 right line maizes on one bide of another, are> 

 together, equal to two right angles. 



Let AB be perpendicular to CD, and 

 EB oblique to it, then f:BE + EBDr: 

 CBA4-ABE-|-EBD=CBA+ABD (14). 



C B 1) 



95. Theorem. If tv^o right lines make 

 with a third, at the same point, but on oppo- 

 site sides, angles together equal to two right 

 angles, they are in the same right line. 



If it be denied, let AB, which together f) 



with AC, makes with AD, the angles y^ \\ 



BAD, DAC equal to two right angles, be p ^ \^ 



not in the right line CAE. Then BAD 

 4-DAC, being equal to two right angles, is equal to EAD 

 -f-DAC (91), and BAD=:EAD, the k-ss to the greater, 

 which is impossible. 



96. Theorem. If two right lines intersect 

 each other, the opposite angles are equal. 



From the equals,ABC-(- ABD and ABD A-^^ ^^1> 



-l-DBE (94, 8-2), subtract ABD, and the 

 remainders, ABC, DBE, are equal. In Q 

 the same manner ABD:::CBE. 



97. Theorem, If one side of a triangle be 

 produced, the exterior angle will be greater 

 than either of the interior opposite angles. 



Bisect AB in C, draw DCE ; take CE A E 



=:CD, and join BE, then the triangle 

 ACD=BCE {c)a, 86), and z.CBE= / /'c^ 

 CAD; but ABF>CBE, therefore AhF> 



D 



BF 



CAD. And in the same manner it may 



be proved, by producing AB, that ABF is greater than 



ADB. 



98. Theorem. The greater side of any 

 triangle is ojjposite to the greater angle. 



LetAB>AC, then ^ACB>ABC. For C 



taking AD=AC,andjoining CD, ^ACD 

 =ADC(87). But aADOCBD (97), A 



and ACB>ACD, therefore much more 

 Z.ACB>CBD, orABC. 



D B 



