43 The Three Sections, Tangencies, 



Again it may be remarked that if we could solve the 

 generating problem by a different method^ we could thence 

 derive other analogovis solutions to the ' the Three Problems 

 of Section/ 



However^ instead of giving another solution to the problem^ 

 in which it is required to have PC.SD : Q,C.E,D : : I : k. 

 I will now solve the more extended generating problem in 

 which it is required to have 



PC.SD +J9.5: Q,C.RD+g.r : : I : /c_, where the magnitudes 

 and signs of the rectangles ja.*, and q.r are given. 



ANALYSIS. 



(The figure to be supplied by the reader.) 



Suppose we draw PG, QG^ making the angle PG and QG 

 right to M^ respectively equals to SN and EN right to B. 

 Then the point G is given. 



If we draw CE and CF meeting PG and QG in E and P, 

 so that the angles EC and FC right to P and Q shall be each 

 equal to angle DB right to R or S. Then it is evident the 

 triangles QCF^, PCE^ are similar to the triangles EBD^ SBD, 

 and that a circle can pass through CFEG. Moreover^ it is 

 evident PC.SD and QC.ED are respectively equal to PE.SB 

 and QF.RB, and therefore we have PE.SB +j9.s : QF.RB + 

 g.r : : I : k. 



And if in PE and QF we take the points J and T^ such 

 that JP.SB and TQ,.RB are equals respectively to j!?.s and g.r; 

 then it is evident the points J and T are knov/n, and that 

 JE.SB : TF.RB : : I : k. Now, fi'om the porisms in rr«W5- 

 actions for 1859, we know that the circle EFG will cut the 

 known circle JTG in a point O such that JO. SB : TO.RB 

 : : I : k ; and therefore O is a known point in circle JTG. 



Again, let V be the point in which CA cuts circle CEFG, 

 and U that in which GV cuts circle JTG, and H that in which 

 UA again cuts circle JTGU. 



We have angle HO right to U or A = GO right to U or 

 V = CO right to V or A, and therefore a circle can pass 

 through COH and A ; but the angle VG right to C being- 

 equal EG right to C it is equal DN right to B, and .*. if W 

 be the point in which GV cuts NN, a circle can pass through 

 VIDW, and the angle WV right to D or N is equal the angle 

 IV or lA right to D or B, and thererefore GW is known in 

 position. Moreover, the point U where GW cuts circle 



