OF SPACE. 



29 



Let ABC and ADC be subtended by AC. Draw 

 the diameter DBE, then /.ABEzrADB-f BAD 

 (108)z=2ADB (87). Also Z.CBE=2CDB; there- 

 fore ABE— CBE=:2ADB—2CDB,orABC=2ADC. 

 In a similar manner it may be proved in other posi- 

 tions. 



134. Theorem. The angle contained by the tangent 

 and any chord, at the point of contact, is equal to the angle 

 contained in the segment on the opposite side of the 

 chord. 



Draw the diameter A B, and join BC ; then L BC A 

 is equal to half the angle subtended at the ceirtre by 

 tlie semicircle AB, or to a right angle, and ABC 

 and BAC make together another right angle (93), 

 therefore deducting BAC, ABCzzCAD. And it 

 appears also from the last proposition, that the angle, li^ A 1> 

 contained in tlic lesser segment CA, is equal to tlie complement of 

 ABC to two right angles, or to CAE. 



135. Problem. To draw a tangent to a circle from a 

 given point without it. 



Join AB, bisect it in C, and on C draw a 

 circle, with tlie radius CB, intersecting the 

 former circle in D, then AD shall touch the 

 circle. For tlio angle ADB, in a semicircle, 

 is a right angle (134, 127), and BD is the ra- 

 dius of the given circle. 



136. Theorem. In equal circles, equal angles stand 

 on equal arcs. 



For the chords of equal angles are fiqual 

 (86), and the segments cut off by them con- 

 tain equal angles (133) ; and if a segment 

 equal to AB be supi)osed to be described on 

 the chord CD, and on the same side with 



CED, it must eohicide with CED, for since, at each point of eocli 

 arc, CD subtends the same angle, the points of one are can never 

 be within those of the other (90) ; the arcs are therefore equal. 



