28 The Three Sections, Tangencies, 



PC and RD shall be to each other in the given ratio of m to n, 

 { — being of given magnitude, and of known sign iti respect to 

 directions on MM and NNyi. 



ANALYSIS. 



Suppose on MM and NN we take PQ and RS, so that 

 PQ : ES : : PC : RD : : m : n, and that we draw PO and 

 QO making- tlie angles PO right to Q, and QO right to P re- 

 spectively equal to the angles RA right to S, and SA right to 

 R. Then it is evident that the triangle POC is similar to 

 RAD, and that the angle OC right to P is equal to the angle 

 AD right to R. Hence, H being the point of intersection of 

 PO and RA, it follows that a circle can pass through AOC 

 and H ; but A, O, and H are knoAvn points : therefore the 

 point C, in which the circle AHO cuts MM, is known, and 

 therefore also the line CAD. 



COMPOSITION. 



On MM and NN take segments PQ and RS, having to 

 each other the given ratio of m : n ; draw PO and QO, 

 making the angles PO right to M, and QO right to ISI equal 

 respectively to the angles RA right to N, and SA right to N ; 

 through A, O, and the intersection H of . PO and AR, de- 

 scribe a circle ; through either point C, in which this circle 

 AHO cuts MM, draw CA to cut NN in D : then will CAD 

 be an answerable line. 



For draw OC. The angle AH or AR right to C or D is equal 

 angle OH or OP right to C, and therefore since the angle 

 PO right to Q or C is equal angle RA right to S or D, it is 

 evident that the triangles POC and RAD are similar, and 

 that PC : RD : : PO : RA : : PQ : RS : : Wi : n. 



DISCUSSION. 



It is evident that when "^ is restricted as to sign, there is 

 but one point O, one circle OAH, and two answerable points 

 C (real or unreal) . 



If - be unrestricted in sign, then, obviously, there are two 

 points O, two corresponding circles AOH, and, therefore, 

 four answerable points C. Moreover, as the points O must 



