134 MR TALBOT ON MALFATTl's PROBLEM. 



The demonstration of this is the same as the last, employing Lemma 7 instead 

 of Lemma 6. These theorems may be called Plucker's tangents, from the name 

 of their discoverer (Crelle's Annals, torn. xi.). It is evident that there are several 

 more cases besides those considered in Lemmas 6, 7, 8, 9; but I omit them, 

 because they are not required for the solution of Malfatti's problem. The 

 demonstration of each would be nearly in the same words. 



Lemma 10. — If three lines issue from a point A, and contain the angles DAX, 

 EAX, which may be called 6 and <P ; and if two circles, with centres B and C, are 



Fig. 7. 



inscribed anywhere in these angles, touching the outer sides at D and E ; then if 

 DE is joined, the intercepted chords DF, GE are in a contant ratio; namely, in 

 the ratio of tan | 6 to tan \<p. 



Demonstration. — Join BD, and draw the perpendicular BH, dividing the 

 chord DF into two equal parts. Draw AI perpendicular to DE. 



Then, since the triangles ADI, BDH are similar, 



AD : AI : : BD : DH 

 . • . DH = AI . ~ = AI . tan £ 

 . • . chord DF = 2 AI . tan \ 6 

 By similar reasoning it may be shown that chord GE = 2AI tan \ $ 



. • . DF : GE : : tan £ 6 : tan \ <p 



Corollary 1.— If 6 = <p t DF = GE. That is :— " If the circles subtend equal 

 angles at A, the intercepted chords are equal." 



