OF SPACK. 



19 



in the sauic manner /. BGCrzBCG ; but BGC > AGC, therefore 

 BGC > ACG; 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 of the point G ; therefore the triangle equal to 

 DEF, supposed to be described on AB, coincides with ABC. 



90. Problem. To bisect a given angle. 



In the rig:lit lines forming the angle, take at pleasure 

 ABizAC ; on BC describe an equilateral triangle B 

 BCD, and AD will bisect the angle BAC. For AB=z 



AC, BDzzCD, and the base AD is common, therefore 

 the triangle ABDizACD (89), and L BADzzCAD. 



91. Problem. To bisect a given right line, AB* 

 Describe on it two eqtiilateral triangles, ABC, 



ABD ; and CD, joining their vertices, will bisect AB 

 in E. For since ACzzCB, ADziBD, and CD is 

 common to the triangles ACD, BCD, /_ ACDzz 

 BCD (89) ; but CE is common to the triangles ACE 

 and BCE, therefore AE=:EB (86). 



92. Problem. To erect a perpendicular to a given 

 right line at a given point. 



On each side of the point A, take at pleasure ABzz 

 AC, and on BC make an equilateral triangle, BCD. 

 Then AD shall be perpendicular to BC For the 

 sides of BAD and CAD are respectively equal, there- 

 fore the angle BADziCAD (89), and both are right 

 angles (64), and AD is perpendicular to BC (65). 



93. Problem. From a point, A, without aright line, 

 BC, to let fall a perpendicular on it. ^ 



On the centre A, through any point D, beyond 

 BC, describe a circle, which must obviously cut BC 

 join AB and AC, and bisect the angle BAC by the 

 line AE; AE will be perpendicular to BG. For 

 Z-BAEzzCAE, ABizAC, and AE is common to 

 the triangles BAE, CAE; therefore Z.AEB=AEC (86), and both 

 are right angles (64). 



C 2 



