Ml 



MATH KM. \TICS PLANE GEOMETRY. 



[BOOK Hi. PEOP i. in. 



8. Prore th convene of 



o. IIUTD ura WIWKTH ui Example 2; namely : If position XI., Book II., and from the greater segment 

 the square* of the sides of a quadrilateral are together cut off a part equal to the less : prove that this greater 

 equal to the squares of the diagonals, the figure must bo ' segment will thus be divided also according to the same 



proposition. 



In Book III., the properties of circles are investigated. 



a parallelogram 

 9. Lot a straight line be divided according to Pro- 



CHAPTER III. 



i:r KMENTS OF EUCLID. BOOK III. 



DBHMT10N8. 



A straight line is said 

 to touch a circle when 

 it mrrtt the circle, and 

 being prolonged, does 

 not cut it 



II. 



Circles are said to touch one another, which meet, but 

 do not cut one another. 



in. 



Straight linos are said to be 

 equally distant from the centre of a 

 circle, when the perpendiculars 

 drawn to them from the centre are 

 equal. 



IV. 



And the straight line on which 

 the greater perpendicular falls is 

 said to be farther from the centre. 



v. 



An arc of a circle is any part of the circumference. 



VI. 



A stgment of a circle is the figure con- 

 tained by an arc of a circle and the 

 straight line joining its extremities, 

 which straight line is called the chord of 

 the arc, or the base of the segment. 



VIL 



An 'angle in a segment is the angle / 

 contained by two straight lines I 

 drawn from any point in the arc of I 

 the segment, to the extremities of \ 

 the chord of the arc, or base of the V 

 segment. Thus A is an angle in \. 

 the segment BCD. ^s 



VIII. 



And an angle is said to insist or stnml upon the arc 

 intercepted between the straight linen that contain the 

 angle. Thus A stands upon the arc B D C. 







A lector of a circle is the figure 

 contained by two straight lines 

 drawn from the centre, and the arc 

 between them. A B C is a lector, 

 so is the remaining part of the circle. 



B 



Similar segments of circles 

 are those in which the angles 

 are equal, or which contain 

 equal angles. 



PROPOSITION I PBOBLBM. 

 To find the centre of a, given circle (A B C). 



Draw any straight line A B, terminating in the cir- 



10 I. cumference, and bisect it in D : * draw D C at 

 + ll I. right angles to AB,t prolong it to E, and bisect 



10 1. C E in F :* the point F shall be the centre of 

 the circle. 



For if it be not, let G, a point out of the line C E, be 

 the centre : and draw G A, G D, G B. Then because 



+ Const. D A = D B, t and D G 

 common to the two triangles A DG, 

 B D G, the two sides A D, D G are 

 = the two B D, D G, each to each, 

 and the base G A = the base G B, 



Def. 12 1. because G is the centre,* 

 . . the angle GDA = GDB:t 



+ 81. but when a straight line 



standing upon another makes the 



adjacent angles equal, each angle is 



. Def. 8 I. a right angle :* . . G D B is a right angle : but 



+ Conrt. F D B is likewise a right angle, t . ' . G D B = 



FDB, the less to the greater, which is impossilili-. . . 



the centre cann to be at any point out of the line C E : it 



must .'. be at a point in CE, and .'.at the middle of 



C E, . . F is the centre, which was to be found. 



COB. From this it is manifest that if a chord of a circle 

 (as A B) be bisected by a perpendicular, that perpen- 

 dicular will pass through the centre. 



PROPOSITION II. THEOREM. 



If two points (A B) be taken in the circumference of a 

 circle (ABC), the straight line or chord (AB) Watch 

 joins them shall fall within the circle. 

 For if not, let it fall without, as A E B. Find D, 



l III. the centre of the circle,* and draw DA, D B. 

 In the arc A B take any point F, draw D F, and produce 



+ Def. 12 1. it to E : then because D A = D B,t the angle 



.51. D AB = DB A:* and be- 

 cause A E, a side of the triangle DAE, 

 is prolonged to B, the exterior angle 



+ 161. DEB is greater than DAE :f 

 but 1) A E = D B E, as just proved : 

 . '. D E B is greater than D B E ; but 

 to the greater angle the greater side is 



19 1 opposite,* . . D B is greater 

 + Dof. 12 1. than D E : but D B = DF,t 



. . D F is greater than D E, which is impossible, . . A B 

 does not fall without the circle. And since DB, the 

 radius of the circle, is greater than D E, and that E is 

 any point on the chord, between A and B, .'. no point 

 between A and B can be on the circumference, .-.the 

 chord AB is wholly within the circle, . . if any two point*, 

 >tc. Q. E. D. 



PROPOSITION III. THIOEKM. 



If a s'raight line (C D) tiirough the centre of a circle 

 (A B C) bisect a chord (A B), which does not pass 

 through the centre, it shall cut it at right angles ; and 

 if it cut the chord at right angles, it shall liisici it. 



l ill. Take E, the centre of the i ircle,* and draw 



