30 Mr. Davies's Symmetrical 



Then, since DF/i L«c, J be are right angles, the linesyD, 

 Lc, eJ are diameters of the circle abF. They also pass 

 through the middles of the arcs ab, b¥, Fa; and are, conse- 

 quently, perpendicular to the middles of the chords ab, bF, Fa, 

 which respectively subtend those arcs. But a bis also a chord 

 of the semicircle AabC; and as yD is a chord perpendicular 

 to the middle of it, it passes through the centre of the semicir- 

 cle, and therefore bisects the diameter AC. Hence D is the 

 middle of AC. 



In the same manner it may be proved that AB, EC are 

 bisected in L and J. 



Prop. III. (fig. 1.) 



Let O be the point of intersection of the perpendiculars 

 Ab, Ca, BF; then the distance of O from each of the angles 

 A, B, C, is bisected by the circle abF. 



For join L(?. Then eL,ab is a quadrilateral in a circle, and 

 the angle ^LA is equal to the opposite angle abA. But 

 aBbO is also a quadrilateral in a circle, and tlierefore the 

 angle ab O is equal to a BO. Hence the angle eLA is equal 

 to ABO, or eL is parallel to BO. Consequently we have 



AL:BO:: A<? : AO:: 1:2; or 

 AO is bisected in the point e. 



In the same manner it appears that^and c are the middles 

 of BO and CO. 



Cor. — Let H be the centre of the circumscribing circle, and 

 the perpendiculars HD, HL, HJ drawn ; we shall have BO 

 equal to twice the perpendicular HD, AO to twice HJ, and 

 CO to twice HL. 



For by the above demonstration Dc, L^ are parallel to CO 

 and BO respectively, and consequently to LH and DH re- 

 spectively: whence <? L is equal to DH. But eL, is half BO, 

 or BO is equal to twice cL or to twice DH. 



The same reasoning applies to the other stated equalities. 



Prop. IV. (fig. 1.) 



Let G be the centre of the circle ab F, H that of the circle 

 ABC, and O the intersection of the perpendiculars; — these 

 three points G, H, O are in one straight line. 



For, since HD is parallel and equal to J'O, the lines y"D, 

 HO bisect each other in their point of intersection, or HO 

 passes through the middle of y"D, the diameter of the circle 

 abF, and therefore through its centre G. 



Cor. 1. — The centre G of the circle a6 F is midway between 

 O and H. 



Cor.2. — It is known (Bland's Geometrical Problems, sectiii. 



Pr. 2*.) 



