MR TALBOT ON MALFATTI S PROBLEM. 137 



This will be seen better by referring to fig. 10, in which, to avoid confusion, I 

 have only represented one of the required circles B', and the two secondary circles 

 a, 7, which belong to it. I have represented the prolongations of the three 

 tangents DN, EN, FN by dotted lines. The secondary circles touch these dotted 

 lines, and also touch the sides of the triangle at D and E, where the tangents 

 intersect them. A simple inspection of the figure suffices to show that the 

 tangent DY drawn from D, which touches both the circles B' and 7, is the sum 

 of two parts, which equal the external tangents DG and EL respectively. And 

 that the tangent EZ, drawn from E, which touches both the circles B' and a, is 

 the sum of two parts, which equal the external tangents EL and DG respectively. 

 Therefore, these two tangents DY, EZ are equal to each other, since each of them 

 equals DG + EL. 



But the three lines DN, EN, FN meet in one point at N ; that is, the three 

 internal common tangents of the circles a, (3, 7 meet in one point. Therefore, by 

 Lemma 8, their other three common tangents must also meet in one point, 

 which point may be called 0. 



Moreover, the line DN produced is the external tangent of the circles B' and 7 ; 

 and EN produced is the external tangent of the same circle B' and circle a ; and 

 (as we said before), FN produced is the internal tangent of the circles a and 7. 

 But DN, EN, FN concur in a point ; that is, two external and one internal common 

 tangents of the circles a, 7, and B' concur in a point. Therefore, by Lemma 9 the 

 other three corresponding tangents of those circles meet in a point. And it is easy 

 to see what that point is. For, the second external common tangent of the circles 

 B' and 7, is AB, one of the sides of the given triangle ; and the second external 

 common tangent of the circles B' and a is BC, one of the sides of the triangle. 



But we know the point of concourse of AB and BC, to be at B, one of the 

 vertices of the triangle. Consequently, we attain this important result, that the 

 second common internal tangent of the circles a and 7 passes through the angular 

 point B of the triangle. In a similar way, it may be shown that the second 

 common internal tangent of the circles a and (3 passes through the angle C ; and 

 that the second common internal tangent of the circles (3 and 7 passes through 

 the angle A. 



These three tangents are therefore three lines proceeding from A, B, C, the 

 three angles of the given triangle, and meeting in a single point (which a few 

 lines previously we named 0). Now Steiner affirms that this point is the 

 centre of the inscribed circle of the given triangle ABC ; and this is the theorem 

 which Plucker and other geometers have been unable to prove except by the 

 use of analytical methods of investigation. But here we call to our assistance 

 the Lemma 11, which we have demonstrated above; and we proceed as follows. 

 Since we have shown that the line BO touches the circle a, and also the circle 7. 

 And since we have also shown that the line DY drawn from the point of contact 



