86 Improvements in Fundamental Ideas 



zero, then it is evident that CB may take any direction 

 through C, and therefore that the point I is that in which any 

 straight line through C cuts the circle CHA : hence in this 

 case the circle IAB coincides with AHC, and the points E 

 are identical with the intersections of the circle AHC and 

 straight line MM. 



Fourthly. — If B and A coincide, then EAD coincides with 

 EBO, and O coincides with D, and circle CAD with circle 

 CBO. In this case the points I and circles IAB are innu- 

 merable, and the problem becomes f porismatic/ and such 

 that any two lines CO BO making the angle OB right to 

 C = 9 right, will cut MM and NN in E and F, so that 

 PE : QF::m:«. 



Fifthly. — It is to be remarked that there are a great num- 

 ber of other particular cases of this problem, arising from 

 particular relative states of the involved data; but, as the 

 solution is general, it holds good in all cases, although it may 

 sometimes require considerable geometrical address to see 

 the modifications necessary. For instance, MM and NN 

 may be coincident, &c. 



DEDUCTIONS. 



(The figure for the following deductions to be supplied 

 by the reader.) 



THEOREMS. 



If P and Q be points in lines MM, NN, and that PB 

 and QC make the angle PB right M equal QC right to N, 

 and that the ratio of PB to QC is as m to n ; then H being 

 the point of intersection of PB and QC, the circle CHB is 

 such that D being in point in its circumference, and EF the 

 points in which DB and CD cut MM and NN, we shall have 

 PE : QF :: PB : QC :: m : n. 



And I being the point of intersection of MM and NN, it 

 is evident a circle can pass through the points IPQH. 



And if O be the other point in which these two circles 

 CHB IPQH cut, it is evident the triangles PBO QCO are 

 similar, and that PO has to QO the same ratio PB has to QC, 

 or which OB has to OC, or of m to n. 



PORISM. 



Given two straight lines MM NN, and two points B C, in 

 position ; then P being any point arbitrarily assumed in MM, 

 a point Q in NN can be found, and an angular magnitude 

 1 0/ such that BD and CD being any two lines through B and 

 C making the angle DB right to C = right, and E and F 



