MR TALBOT ON MALFATTl'S PROBLEM. 135 



Corollary 2. — If 6 is greater than </>, then DF is greater than GE. 



Lemma 11. — If any angle DAE is bisected by the line AX, and two circles 

 are inscribed anywhere in the semi-angles DAX, EAX touching the sides at D 

 and E ; then the tangent DK is equal to the tangent EL. 



Demonstration. — Join DE. We have shown in Lemma 10 that in this case the 



Fig. 8. 



intercepted chords DF, GE are equal. Subtract them from the whole line DE, 

 and the remainders DG, EF will be equal. Therefore 



DG . DE = EF . ED . \ DK 2 = EL 2 

 and 



.-. DK = EL 



Corollary. — Conversely, if DK = EL it follows that the angle DAX = angle 

 EAX. 



For, if those angles are not equal, let DAX be the greater. Then because 

 the angle DAX is greater than the angle EAX, the chord DF is greater than the 

 chord EG (by Lemma 10, corollary 2). Subtract them successively from the 

 line DE, and the remainder EF will be smaller than the remainder DG, therefore 

 EF . ED is less than GE . DE ; therefore EL 2 is less than DK 2 ; and EL is less 

 than DK. But on the contrary EL = DK by hypothesis. Consequently it is not 

 true that the angle DAX is greater than the angle EAX. In the same way it 

 is shown that it is not less ; consequently it is equal to EAX. Q.E.D. 



Lemmas 10 and 1 1 are particular cases of a much more general theorem which 

 I propose to give on another occasion. In the first nine Lemmas I have chiefly 

 followed Plucker, but have endeavoured to make his demonstrations more 

 rigorous by going more into detail than he has done. But Lemmas 10 and 11, 



VOL. XXIV. PART. I. 2 P 



