and Elementary Theorems of Geometry. 89 



points in ivhich OB and OC respectively cut MM and NN, 

 toe shall have the rectangle under PE and QF egwa/ to a given 

 rectangle m . n. 



ANALYSIS. 



Suppose we draw PA making angle PA right to E equal 

 QF, right to C, and PA . QC equal in sign and magnitude to 

 PE . QF : then it is evident the triangle APE is similar to 

 FQC, and that angle EP left to A is equal CQ left to F; 

 therefore if H be the point in which AH parallel to MM cuts 

 QC, and D the point in which EA cuts FC, it follows that a 

 circle can pass through ACDH. It is also evident that the 

 points AH and circle AHC are given. 



And since the angle OB right to C is given in magnitude, 

 the circle OBC is given ; and therefore also the point I in 

 which is again cuts circle AHC. Again, the angle AI right 

 to E or D being equal to angle CI right to D or O, it is 

 equal angle BI right to O or E ; therefore a circle can pass 

 through AIBE. But AI and B are given points ; thsre- 

 fore the circle AIB is given, and hence the point E where 

 it cuts MM, and therefore EBO and COF. 



COMPOSITION. 



Draw QC ; through P draw PA making angle PA right to 

 M equal QN right to C, and on it take PA so that PA.QC 

 shall be equal in sign and magnitude to m n ; through CA 

 and the point H in which AH parallel to MM cuts QC de- 

 scribe a circle ; through B and C describe a circle whose cir- 

 cumference is such that BB and CC being any two lines 

 through B and C to cut on it, we shall have angle BB right 

 to CC equal angle right ; through the points AB, and the 

 other point I in which this circle cuts circle AHC describe a 

 circle ; through either point E in which circle ABI cuts 

 MM, draw EB to cut circle CIB in O ; draw OC to cut NN 

 in F : then will EBO and DCO be as required. 



For let D be the intersection of EA and FC. 



The angle AE or AD right to I equals angle BE or BO 

 right to I = equal CO or CD right to I, therefore the point D 

 lies in circumference AHC. 



The angle AD right to H or EA right to P = CD or CF 

 right to H or Q ; and, therefore, since angle PE right to 

 A = QC right to F, the triangles PAE, QCF, give us PEQF 

 = PA-QC = m-n. 



DISCUSSION. 



If the signs of m and n be given, and that the directions 



M 



