MR TALBOT ON MALFATTI S PROBLEM. 



133 



Lemma 7. — Let A, B, C, be three circles. Let DE be the external tangent 

 of A and B. Let FG be the external tangent of A and C ; and let HI be the 

 internal tangent of B and C. Then if these 

 three tangents, when produced, meet in a single 

 point 0, 



FG - DE = HI 

 For, 



FG - DE = OG - OE = 01 - OH = HI. 



Lemma 8. — If three internal common tan- 

 gents of three circles meet in a point 0, their 

 other three internal common tangents meet in 

 another point P. 



Demonstration. — In figure 5, suppose the 

 common tangents meeting in to be effaced, 

 and replaced by the three other internal com- 

 mon tangents, it is required to show that these 

 also meet in a point. 



Let two of them, viz. those touching the Fig. 6. 



circle A, meet in the point P. Denoting the new tangents by the same letters 

 as before, but accentuated, we have of course 



DE = DE, FG = FG, HI = HI 

 Now we proved in Lemma 6 that 



DE = FG + HI, or ED - HI = FG 

 therefore we have 



ED - Hi = FG 



The new common tangents meeting in P will be PED, PHI ; but these have 

 a part which is equal in each, namely PE' = PH', which are tangents to the same 

 circle A. Therefore subtracting this part, we have 



PED - PHI = ED - HI 



which we proved to be equal to FG. 



Therefore P is a point from which tangents PD', PI have been drawn to the 

 circles B and C, and their difference has been found equal to FG the internal 

 common tangent of B and C. Therefore by the corollary to Lemma 5, P is a point 

 in F(jt produced. Q. E. D. 



Lemma 9. — If two external and one internal common tangents of three circles 

 meet in a point O, the three other corresponding tangents meet in another 

 point P. 



