58 The Three Sections, Tangencies, 



Since tlie rectangle under the half sum and half difference 

 of the sides of a triangle is equal to the rectangle under the 

 half sum and half difference of the segments of the base made 

 by a perpendicular from the vertex ; .'.it is evident^ that if 

 OP be perpendicular to AB^ and M be middle point of AB, 

 and that in OB we have Oe =: OA^ and b the middle of eB^ 

 then will Ob.V>b = AM.MP ; and hence Ob has to PM the 

 same known ratio which AM has to 6B. 



Now if in AB we find the point G such that eb is to GM 

 in the known ratio of Ob to VW, then will Oe or its equal 

 AO have to PG the same kno^vn ratio. 



For like reasons it is e^ddent that if OQ be perpendicular 

 to AC^ and N the middle point of AC^ and that we assume 

 0/ in OC and = to OA, and that c is middle offC, and that 

 we find H in AC such that fc shall have to HN the known 

 ratio of AN to Cc, then will AO have to CIH this same known 

 ratio. 



Now the points G and H are known^ and the ratio of PG 

 to QH is known (because AO has known ratios to PG and 

 QH) : hence the point I in which the circle AHG cuts the 

 circle QAP is known; and since the angles P and Q are right, 

 it follows that the straight line 10 perpendicular to AI is 

 known_, as also the point U in which it again cuts the circle 

 AGH. 



Again, GE being perpendicular to AG, it is parallel to PO; 

 and PG has to OR a known ratio; therefore AO has to OR 

 a known ratio. 



If S he the point in which AO again cuts the circle AGH, 

 it follows, from similar triangles, that AI has to RS the 

 known ratio which AO has to OR, and .*. the chord RS is of 

 known magnitude, and .•. also it is known in position; and 

 AS is known, and also the point O where it cuts IR, and .-. 

 the required circle is kno"v\ai. 



COMPOSITION, 



Through the centres B and C draw BE' and CF' any two 

 radii of the circles B and C; from E' in either direction on 

 EB make E'e' = radius of circle A; from F' in either direc- 

 tion on F'C make F'/' = radius of circle A; bisect e'B in b', 

 Of in c', AB in M, and AC in N; find G in AB such that 

 e'b' : GM : : AM : e'b'; and in AC find the point H such 

 that/'c' : HN : : AN : fc' ; assume any straight line x and 

 find lines y and z such that x : y : : AM : e'b' , and x : z 



