126 GEOMETRICAL PORISMS. 



a ftralght line through P, a given point, without them, t«, 

 meet them in D and E, fo that the redangle BD, CE 

 may be equal to a given fpace. 



Suppose that DE is drawn as required. Join PC which will 

 be given in pofition and magnitude. Draw PF parallel to AC, 

 and take F, fo that the redlangle CP, PF may be equal to the 

 given fpace, the point F will therefore be given ; draw FL pa- 

 rallel to CP, meeting AB in K, and PD in L, then FL and the 

 point K will both be given by pofition. The triangles LFP, 

 PCE are fimilar ; therefore LF is to FP as PC to CE, and the 

 redtangle LF, CE is equal to the redangle FP, PC, which, by 

 hypothecs, is equal to the redangle BD, CE, therefore FL is 

 equal to BD ; now, B and F are given points, and BK, FK are 

 lines given by pofition ; therefore (Prop, i.) if a circle be de- 

 fcribed through K, B, F, there is a given point H in the circum.- 

 fei-ence, fuch, that K, H, L, D are in a circle ; therefore, if this 

 point be found, and HD, HL, HK joined, the angle HDL is 

 equal to HKL; tlierefore HDP is equal to HKF, that is, to a 

 given angle ; but H and P are given points, therefore D is in 

 the circimiference of a given circle ; but it is alfo in a ftraight 

 line given by pofition ; therefore D is a given point, and PD is 

 given by pofition. 



Construction. Join P and C, either of the given points in 

 the given lines ; draw PF parallel to CA, and take F, fo that the 

 given fpace may be the redangle CP, PF. Draw FL parallel to 

 CP, meeting AB in K, and through the points F, B, K defcribe 

 a circle. Find H in the circumference, fo that BH may be 

 equal to FH. Join HK and HP, upon which defcribe a feg- 

 ment of a circle, that may contain an angle equal to HKF ; this 

 circle may meet AB in two points D, 3. Join PD and P5, meet- 

 ing 



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