Ííflf 
THE CONSTRUCTION OF 
gles QGE, QBA, &c. will be equal, and likewise 
SF : FG :: Lill : mn\ but Lni (by construction) is- 
( — lip) = rib, and mn ( — pc/). — hr; therefore SF 
: FG :: db : fee, N and consequently EF (2SF) : FO 
:: ab (2bd) : be. Q. E. 1). 
METHOD OF CALCULATION. 
Ln (dc) : Lm. (db) :: the tangent of LQ/i (the com 
plement of the given angle QBIl); the tangent of LQm; 
therefore, in the triangle OQF, will be given one angle 
OQF and two sides, QO, FO; whence, not only the 
angle SOF, but also SO and SF will be known. 
PROBLEM LI. 
To apply, or inscribe, a given right-line AD between 
the peripheries of two circles C and O, given in magnitude 
and position, so as to be inclined to the right-line CO 
joining the centres in a given angle. 
CONSTRUCTION. 
Make OCB equal to the given angle, and let CB be 
taken equal to the given line ; upon the céntre B, with 
the radius of the circle C, let the arch /¿Dm be described, 
/CU 
rutting the circle O in D ; then draw BD, and parallel 
thereto, draw CA, meeting the periphery in A; join 
A, D. and the thing is done. 
DEMONSTRATION. 
Because (by construction) CA and BD are equal and 
parallel, therefore will AD and CB be also equal and 
parallel (by Euc. 33. l.). Q. E. D.