MB TALBOT ON MALFATTl's PBOBLEM. 



D to the circle 7 , is equal to the line EZ drawn from the point of contact E to 

 ton h the ctrcle a. Therefore by the corollary to Lemma 11. the line BO neces- 

 sarily b.sects the angle B of the original triangle. Similarly it is shown that 

 CO Insects the angle C, and that AO bisects the angle A. Therefore oTt h 

 centre of the mscnbed circle of the triangle ABC : -which is Stub's Theorem 

 The solution of Malfatti's Problem is therefore as follows — 



Bisect the angles of the triangle ABC, by the lines AO, BO, CO In 

 two of the smaller triangles thus made AOB, BOC, inscribe the circles 7 and « 

 From D, the pomt of contact of „ with the side BC, draw a line DY, touchin" the 

 circle y. Then DY will touch one of the required circles also; which c ^0 

 touches AB, BC, two sides of the triangle, and is therefore whollv deemed 



