236 PHILOSOPHICAL TRANSACTIONS. [aNNO 1685. 



Construction. — Take any line a e at pleasure, fig. 13, and at its extremities 

 make the angles e a c, eab, aec, a e d, equal to the corresponding observed 

 angles E A C, EAB, A E C, A E D. Produce b a, de, till they meet in /, and 

 join c/; then upon CB describe (according to 33, 3, Eucl.) a segment of a 

 circle, that may contain an angle = cfb : and upon C D describe a segment of 

 a circle capable of an angle = cfd; suppose F tlie common section of these 

 2 circles; join FB, FC, FD ; then from the point C, draw the lines CA, CE, 

 so that the angle F C A may be = fc a, and FCE = fee; then A, E, the 

 common sections of CA, CE, with FB, FD, will be the points required, 

 from whence the rest is easily deduced. 



Calculation. Assuming a e of any number of parts; in the triangles a c e, 

 a/e, all the angles being given, with the side a e assumed, the sides ac, ec^ 

 af, ef, will be found; then in the triangle caf, the angle oaf with the 

 legs a c, af being known, the angles afc, a cf with the side fc will be 

 known : then for the rest of the work in the other figure, the triangle BCD 

 having all its sides and angles known, and the angles B F C, BFD, being equal 

 to the found bfc, bfd; how to find FB, FC, FD by calculation, and also 

 protraction, has been shown by Mr, Collins, in Phil. Transactions, N" 69, p. 

 563, vol. ], as to all its cases. 



But here it must be noted, that if the sum of the observed angles, B A E, 

 A E D, be 1 80 degrees, then A B and E D cannot meet, because they are 

 parallel, and consequently the given solution cannot take place ; for which 

 reason another is here subjoined. 



Another Solution. — Upon B C, fig. 14, describe a segment B A C of a circle, 

 so that the angle of the segment may be equal to the observed angle b ac^ 

 (which, as above quoted, is shown 33, 3, Euclid) ; and upon C D describe a 

 segment C E D of a circle, capable of an angle equal to the observed C E D ; 

 from C draw the diameters of these circles C G, C H ; then upon C G describe 

 a segment of a circle G F C, capable of an angle equal to the observed angle 

 AEC; likewise upon C H describe a circle's segment CFH, capable of an 

 angle equal to the observed C A E : suppose F the common section of the two 

 last circles HFC, G F C, join F H, cutting the circle H E C in E ; join also 

 F G, cutting the circle G A C in A : then A, E, are the points required. 



Demonstration. — For the angle BAC is = ^ac by construction of the seg- 

 ment, also the angles C E H, C A G, are right, because each is in a semi- 

 circle : therefore a circle being described on C F as a diameter, will pass through 

 £, A. Therefore the angle CAE = CFE = CFH=(by construction) to 

 the observed angle cae. In like manner theZ.CEA=:CFA=CFG;?= 

 the observed angle cea. 



