THE CONSTRUCTION OF 
evidentthat the points, F and G must coincide, and that 
.aE, BG (BE) and CH will represent the true position 
of the laves: suppose KG, KH, PG, PH, PE, and 
PF to be now drawn ; then, since [by construction) GI 
r: HI, and the angle GIK — HIK, therefore is GK 
rr HK (Euc. 4. 1.) : moreover, since KP is [by con 
struction) perpendicular to BC, it will also be p< rpen- 
dicular to the plane BCHG, and consequently the 
augles PKG and PKH botii ritrht angles: therefore, see 
ing the two triangles GKP, HKP have two sides and an 
included angle equal, the remaining sides PG and PH 
must likewise be equal [Euc. 4. j.). After the very 
same manner it is proved that PF (or PG) is equal Uf 
£P. Q. E. D. 
METHOD OF CALCULATION. 
Draw Ir perpendicular, and Hq parallel to BC; then, 
by reason of the similar triangles, Hr/G andlrK, it will 
be as BC (IIq) : BG — CH (G ? ) :: BG -+.. CH (Ir) 
5» 
: Kr — 
BG —CH x BG f CH 
2BC " ' 
which subtracted 
from Br ( = f BC ) gives BK : and in the same man 
ner will BN be found; then in the trapezium KBNP 
will be given all the angles and the two sides BK and 
BN ; from whence the remaining sides, &c. may he 
easily determined. 
PROBLEM LXXVI. 
The base, the perpendicular, and the difference of the 
sides beinggiven t to determine the triangle. 
CONSTRUCTION. 
Bisect the base AB in C, and in it take CD a third 
proportional to 2AB and the given ditlerence of the 
sides MX; erect DE equal to the given perpendicu 
lar, and draw EK parallel to AB, and take therein EF 
— MX; draw EAG, to which, from F, apply FG rz 
AB; draw AH parallel to FG meeting EK in II j 
then draw BH, and the thing is done.