CHAMBERS'S INFORMATION FOR THE PEOPLE. 



chord becomes the point of contact, and the 

 circles touch each other externally. 



Again, if the circle O remain fixed, and the 

 circle Q be supposed to be 

 drawn inwards, the points of 

 intersection A and B will ulti- 

 mately coincide The com- 

 mon chord will become the 

 common tangent, and the 

 middle point of this chord 

 the point of contact, and the 

 circles will touch each other 

 internally. Hence, if two 

 circles touch each other either internally or exter- 

 nally, the line which joins their centres passes 

 through the point of contact. 



Arcs and Angles. 



PROP. XV. The angle at the centre of a circle 

 is double of the angle at the circumference standing 

 on the same arc. 



Let AOB (fig. i) be an angle at the centre of 

 the circle standing upon the arc AB, and let ACB 

 be an angle at the circumfer- 

 ence standing on the same 

 arc ; it is required to prove 

 that the angle AOB is double 

 of the angle ACB. Join CO, 

 and produce it to S. Then 

 in the triangle OCA we have 

 the exterior angle AOS equal 

 to the two angles OCA, OAC ; 

 but these angles are equal, 

 therefore the exterior angle 

 AOS is double of the angle ACO. 

 exterior angle BOS of the 

 triangle BOC is double of 

 the angle BCO, therefore the 

 whole angle AOB at the 

 centre is double of the whole 

 angle ACB. 



Similarly (fig. 2), it can be 

 shewn that the exterior angle 

 SOB is double of the angle 

 OCB, and the part SOA is 

 double of the part OCA ; 

 therefore the remaining angle AOB is double of 

 the remaining angle ACB. 



It follows from this (i.) 

 That the angles ACB, ADB, 

 AEB, &c. at the circumfer- 

 ence standing on the same 

 arc AB are equal to each 

 other ; for each of these angles 

 is half of the angle AOB at 

 the centre on the same arc. 



(2.) That the angle ACB at 

 the circumference, which has a semicircle for its 

 arc, is a right angle ; for if 

 the arc AB be a semicircle, 

 AO and OB are in one line, 

 and the angle AOB at the 

 centre is an angle of two right 

 angles ; therefore the angle 

 ACB is one right angle. 



(3.) That the opposite angles 

 ABC, ADC of a four-sided 

 figure inscribed in the circle 

 are together equal to two right angles ; for, 



618 



Fig. i. 



Similarly, the 



Fig. 2. 



drawing the diameter AE, it 

 is evident that the angle ABC 

 falls short of a right angle by 

 the angle EEC, and that the 

 angle ADC exceeds a right 

 angle by the angle EDC ; but 

 EEC and EDC are equal, 

 being angles at the circumfer- 

 ence on the same arc EC, 

 therefore the angle ABC falls 

 short of a right angle by as much as the angle 

 ADC exceeds the right angle ; therefore the two 

 angles ABC, ADC together are equal to two right 

 angles. 



(4.) That the angle ABD, 

 between the chord AB and 

 the tangent BD at the point 

 B, is equal to any angle E 

 in the alternate segment. 

 For the angles AEB, AFB, 

 &c. are all equal ; and when 

 the point E approaches 

 B, and ultimately coincides 

 with it, AE becomes AB, 

 and EB becomes the tangent BD, and the angle 

 at E becomes the angle ABD. 



PROP. XVI. Equal arcs have equal chords, 

 and conversely. 



Let AB and CD be two equal arcs ; it is re- 

 quired to prove that the chord 

 AB is equal to the chord CD. 

 Since the arc AB is equal, to 

 the arc CD, the angle AOB is A , 

 equal to the angle COD, for 

 the arc is the measure of the 

 angle. Then comparing the 

 two triangles AOB, COD we 

 have two sides of the one equal 

 to two sides of the other, and the contained angles 

 equal ; therefore the base AB is equal to the base 

 CD. The converse proposition is evident. 



CONSTRUCTIONS (IV.). 



(i.) To describe a circle touching a given lint 

 in a given point, and passing through a given 

 point. Let AB be the given 

 line, C the given point in it, 

 and D the point through 

 which the circle passes. 

 From C draw a line CS 

 perpendicular to AB the 

 centre of the circle lies in 

 this line. Join CD. Bisect 

 it in E, and through E draw 

 EO perpendicular to CD 

 the centre of the circle is in this line also ; there- 

 fore it must be at O their 

 common point. Hence, 

 from O as centre, with 

 radius equal to OC, describe 

 a circle. This circle passes 

 through C and D, and AB 

 touches it. 



(2.) On a given line to 

 describe a segment of a 

 circle containing an angle 

 equal to a given angle. Let 

 A be the given angle, BC 

 the given line ; it is required 

 to describe on BC a seg- 



