88 



Improvements in Fundamental Ideas 



Let HQ be fixed points, and H any point in a fixed circle ; 

 a point O can be found, such that any circle through it and the 

 point H shall cut PH and QH in B and C, so that D and I 

 being any two points in the respective circles, BHC and BHQ, 

 and EF the points in which DB and DC cut IP IQ, then will 

 PB have to QC and PE have to QF the given ratio which m 

 has to n. 



The point O is evidently in circle PHQ, and PO : QC :: m : 

 n, &c. 



PORTSM. 



Given the points P and Q in given straight lines MM NN ; 

 a point C and angle 6 right can be found, such that E and F 

 being any two points in the given lines, such that CE right to 

 F shall be equal angle right, then will PE have to Q,F a 

 given ratio of m to n. 



If we describe a circle through PQ, and the intersection of 

 the given lines, the point C is evidently in its circumference, 

 and such that PC : QC ::m:n. The 6 angle is evidently equal 

 MM right to NN. 



Given two straight lines MM NN, and the point P in MM, 

 and Q in NN ; through two given 'points BC, to draw two 

 straight lines BO CO, making an angle OB right to C of a 

 given magnitude right, and such that E and F being the 



