and Elementary Theorems of Geometry. 83 



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

 P in MM, and Q in NN ; through two given points BC to 

 draw two lines BO CO, making the angle OB right to C equal 

 a given angular magnitude right, and such that E and F 

 being the respective points in which OB and OC cut MM and 

 NN, ?#e shall have PE fo QR in the given ratio m to n. 



ANALYSIS. 



Draw QC. Suppose we draw PA making angle PA right to 

 E = angle QC right to F, and PA : QC : : PE to QF ; then 

 the triangle APE is evidently similar to CQF, and angle AP 

 right to E = angle CQ right to F, and therefore, H and D 

 being the points in which PA and EA cut QC and FC, a 

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

 the points AH and circle AHC are given. Since angle OB 

 right to C is given in magnitude, and B and C in position, the 

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

 it again cuts circle AHC. Again, the angle AI right to E or 

 D being equal to the 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 AIB are given points ; hence this circle is given, 

 and the point E in which it cuts MM, and therefore also the 

 lines EBO OCF. 



COMPOSITION. 



Draw QC ; through P draw PA making angle PA right 

 to M = QC right to N, and on it take PA, such that 

 PA : QC :: m : n; through C A and the point II in which 

 PA cuts QC, describe a circle ; through B and C describe 

 that circle such that an angle at any point of its circumfe- 

 rence, from B right to C, shall be equal angle right ; 

 through the points A B 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, FCO, 

 be as required. For let D be the point of intersection of 

 EA and FC. 



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

 to I = CO or CD right to I ; and therefore the point D is 

 in circumference AHC. The angle AD or AE right to H or 

 P = CD or CF right to H orQ; and since PA right to 

 E = to QC right to F, the triangles PAE QCF give as 

 PE : QF :: PA : QC :: m : n. 



DISCUSSION. 



If the signs of the directions on MM and NN are known, 



