New Principles. of Gardening. 7 
Practice. Makedé equalto z4 and #0, make @¢ equal to Fig. XU. 
no; and ftom, raife the Perpendicular ¢4. Bilect dé in 
a, and on 4, with the Interval ¢ d, or 24, defcribe the Semicir- 
cle dh, cutting the Perpendicular in 4; 4¢ will be the mean 
Proportion. For, (44, 44, being joined) the Angle 4d is 
right-angled. Therefore be is. the mean Proportion between 
cbandcd QE... 
PROBLEM XI. 
30 divide a Right Line (bf) in extream and mean Pre- 
portion. 
Dermuition. A Right Line is faid to be divided ta extream 
and mean Proportion, when the Wholeis to the greateft Segment, 
as the greatelt Segment is to the Leffer. 
Practice. Make wv equal to Of, and bifet wv by 
xh,inr ; throughband x drawst and parallel to wv; and 
through w and wv the Lines ¢ y, and s% parallel to bx; 
on t, with the Interval. ty, defcribe the Arch 9 #75, cutting 
hx in #, andwvino. Draw 04, and on 4 with the Inter- 
val bo, cut 4x ina, fo fhall ba be the greater Segment, and 
ax the lefler; and confequently 4x is thereby divided in ex- 
tream and mean Proportion. Let x0 be drawn, and with that 
Interval on x, defcribe the Arch ac, and on 4 the Arch 0 a, and 
draw the Right Lines 0@ and oc; then will the Angle. heobe 
equal to the Angles ¢ xo, and xoc; becaufe thefe two Angles 
are within, the other without the Triangle ¢xo. Likewife 
the fame Angle 4co is equal to the Angles aac and aoc 
and oca equal to oc; alfo the Angles ox¢ and ¢oa@ are 
equal toeach other. Therefore the Triangles x o¢, andaa¢, 
have two Angles equal to two Angles of the Triangle ¢oa, 
and therefore equal-angled. For as ox or oh is toc, fo is 
oc, or oa, to ac. ‘Therefore oc or o@ is a mean Propor- Fig, XII: 
tional between bo and ¢a. On the Right Lines ox, 04, make 
hd and xe, equal to 4c, and join de and de; then will do 
and oe be equal, and de parallel to 4, and confequently the 
alternate Angles dea, xae, and eda, hed, ave all equal 3 
, e€ac 
