386 



THE NINE-POINTS CIRCLE. 



Obviously the only way of taking a piece off l^p and adding 

 it to dl and still retaining the proportion is to reverse the figure 

 on the diagonal of the square. 



It is equally obvious that x lies on opposite sides of the 

 mid-point of dp in the two positions. If, therefore, we have 

 occasion to compare a range of points DLXP with D/.rP, know- 

 ing that /;r=LX, and have given that Dl.DF==Dx^ and also 

 DL.DP=DX2, we know that / coincides with L and x with X, if 

 only we know that x and X are on the same side of the mid-point 

 of DP. 



Perhaps I might have assumed this elementary truth. 



Relation of tzvo circles in fangency. — If two circles touch, 

 and a tangent be drawn to one of them meeting the other as a 

 chord; if then the ends of the chord be joined to the contact 

 point of the circles, and these joining lines mark off a new chord 

 in the second circle ; if also, at the end of this new chord, tangents 



DL:Lm::nU:UF 

 i.e., DL : LX : : XL^ : L^P. 

 Or, as we have seen 

 above, DL.DP=DX2. 



Fig. 3- 



be drawn to meet the original chord, then that chord is divided 

 in the proportion we have just been studying. When the figure 

 is drawn, the proof is immediately obvious from the similarity 

 of the triangles DLw and nUF. 



These two very simple propositions are here set down merely 

 for convenience. They are both of them interesting enough 

 to have been taught long before students come to the associated 

 circles. They do not, therefore, require proof from me, but 

 only enunciation. But when they are enounced, the truth of 

 Feuerbach's theorem becomes almost axiomatic. 



Proof of Feuerbach's theorem : — • 



Given the triangle ABC with its Link circle and the in-circle. 



Slide a circle equal to the in-circle along the chord DP until 

 it touches the Link circle on the side nearer P. Join the point 

 of contact with D, and where it cuts the inner circle draw a 

 tangent, cutting DP in /. If the moving circle touches DP at x, 

 we have D/.DPr=D,r2. 



