GEOMETRICAL PROBLEMS. 
38i 
mi x EL, and consequently that the triangles ÈHL 
and EAG are also equal to each other (Eue. 15.6.) 
irom which taking away EDC, common, the remain 
ders CÜHL and CDGA will be equal likewise, and 
consequently ALHB — AGB, being the differences 
between those remainders and ACDB. But the triangle 
ADF is rr ACD, standing upon the same base AD 
and between the same parallels ; therefore, (by adding 
AGD, common) AGF is also r= CDGA ( — CDHL) ; 
but AGF (CDHL) : AGB (ALHB) :: GF : GB 
(Eue. 1.6.). Q.E.D. 
METHOD OF CALCULATION. 
In the triangles ABE and ABK are given all the 
angles and the side AB, whence BE, BK, and EC will 
be known; then, in the triangle EEC will be given 
all the angles and the side CE, whence EF, and from 
thence FG, and EG, will be known ; lastly, from the 
known values of EK, EG, and EF, the value of FH 
( — \/EG x EK — EF will be found. 
PROBLEM LXVIII, 
Two right-lines AG and AH, meeting in a point A, 
being given by position; it is required to drain a right- 
line n\ 3 to cut those lines in given angles, so that the 
triangle AnP, formed from thence, may be equal to a 
given square ABCD. 
CONSTRUCTION. 
Let the angle ABE be equal to the given angle APn 
and let BE 
meet AG in 
' E; draw EF 
make BQ 
equal 2FIF, 
and upon 
AQ describe 
A F B SQ PH 
the semi-circle A wiQ, cutting BC in m; draw mn pa 
rallel to AH, meeting AG in n, and nP parallel to EB, 
and A»P will be the triangle required.