438 
GEOMETRY. 
terras, and inchide a fpace or figure, which is abfurd. 
Therefore DEB and D F B are not right lines j neither 
can any otlier lines drawn from D to B, befides B D, he 
right lines. Confequently the line D B, the common 
fecbion of the planes, is a right line. 
310 . If two planes A B, CD, which are both perpendicular 
to a third plane E F, cut one another ; 
their interjeElion H G is at right angles 
to that third plane E F.—For the com¬ 
mon fedtion of A B and C D is a 
right line GH (309). Alfo H B, 
H D are the common fection's of 
AB, CD, witli the plane EF. Now 
from the point H, a line HG drawn 
perpendicular to the plane E F, 
mud beat right angles to H B, HD (305).. But FIG 
miilt be in both planes AB, CD (208). Therefore it 
mult be in the common fettion of thole planes. Confe- 
quently the fedlion HG of the planes A B, CD, is at 
right angles to the plane EF. 
Methods of determining Distances by means of 
limiiar Triangles traced on the Ground. 
211. To find the length of the line AB acceffible only at 
both ends. —•Having fixed on fome 
convenient point P, meafure BP 
Rnd AP; and prolong thole lines 
till PGrrtPB, and PD=PA ; then 
the diftance between the points D 
and G will be equal to AB. For 
the fides of the triangles GPD, 
BPA, about the equal angles at P, 
are refpettively equal, therefore the 
third fides GD, BA, will alfo be 
equal. 
Or thus:—Having meafured PB, PA, (as before,) 
take PC fome convenient aliquot part of PB, and PO 
the fame aliquot part of PA ; then meafure the crofs 
diftance OC, which will be the like aliquot part of the 
required diftance AB. For the fides PO, PA ; PC, PB, 
being proportional, the triangles OPC, APB, will be 
limiiar; hence PO : PA :: 0 "C : AB ; therefore what¬ 
ever part PO is of PA, the like OC will be of AB. 
Suppofe PAr2392, and PB~4i4 feet; and let PO, 
PC, be i of PA, PB, or equal to ySf. 5in. and Szf. 9|in. 
And fuppofe OC meafures 93A feet; then ABzr93§X5 
=4674 feet. 
312 . When the line (AB) is acceffible at one end (B) only. 
^ . —We luppofe fome object at 
< 3 - PA. inaccefiible end A; and 
T let a mark be fet up at B ; 
\ then in the direction AB take 
BG (the longer the better), 
(P and through a convenient point 
P, as in the preceding pro¬ 
blem, let the diftances BD, 
GR, be meafured fo that PD 
1 C D R- =PB, and PR=2:PG; then if 
a mark be fet up at K, the interfeClion of AP, and RD, 
when produced, DK will be equal to AB. For the 
triangles PBG, PDR, being fimilar and equal in all re- 
fpeCts, the triangles PBA, PDK, will alfo be fimilar 
and equal. Or, BA may be found without the diftances 
PD, PR, thus: take PC, PO, like aliquot parts of 
PG, PB; then SO will be the fame aliquot part of 
BA. For PO : PB :: OS ; BA. 
Suppofe PB=442, PG=464 feet; and that P 0 = 
jioF, PC=;ii6 feet (^ of PB and PG); alfo, fuppofe 
OS meafures 113 feet; then BA=4.53 fcetj for iio| ; 
442 113 : 452. 
213. Let O be an objebl On the oppbjile fide of a riverto 
find the difiance DO.—Lay down 
an ifofceles triangle DBA, the 
fide DB being in any conve- Or. 
nient direction; tlien having 
meafured a bafe DR, fet up a 
mark at R; and in the fame 
direction take another bafe Rrf, 
and make the triangle dba fi¬ 
milar and equal to DBA {da 
being parallel and equal to 
DA); tlien find the concourfe 
(C) of the lines ORC, daC, 
D 
A. 
and meafure 
RD : DO 
Suppofe DRr 
80 : 273! : : 300 
d & 
w 
'i 
c 
dC. By fimilar triangles, Rd : dC : ; 
1300, R(f=go, and dC-z-]-!,^ feet, then 
: 1014 feet nearly — DO. 
214. But the moft expeditious method of finding the 
diftance to an inaccellible objeCt, is by - 
means of a rhombus, as follows: 
Suppofe O the objcbl, and OB the required dif- Q f 
tance. —With a line or meafuring tape, whole 
length is equal to the fide of tlie intended 
rhombus, lay down one lide BA in the di- i 
reCftion BO, and let BC, anotlier fide, he in 
any convenient direCbion: fallen two ends 
of two of tliofe lines at C and A, then the 
other ends (at D) being kept together, and B C 
the lines llretched on the ground, thole lines AD, CD, 
will form the other two fides of the rhombus. Set up 
a mark at R where CO, AD, interfeCl; and meafure 
RD. Then the fides of the triangles RDC, CBO, being 
refpeCtively parallel, the triangles will be limiiar; 
hence, RD : DC 1: CB : BO. Suppole tlie fide of the 
rhombus is lOO feet, and RDrri if. 7in. then ii^ ; 100 :j 
100 : 863 feet nearly = BO. 
Or thus : Having laid down the rhom- Yi)* 
bus, mark the concourfe of tlie lines '^j\ 
ODS, BCS, and meafure CS ; then CS : s==L\ 
CD :: AD : AO. 
If tlie ground be nearly level, a rhom. | 
bus whole lide is 100 feet, will deter- ^ 
mine diftances to the extent of 300 
yards, within a very few feet of tlie 
truth. 
D 
B 
C S 
'homhus .—At 
mt point B, JjY . ■n* 
1(3 i-lirtinKiic . YV-Vk. 
For 
equal 
215. To find the length of an inacceffible line (QR) by^ 
means of a rhombus. —At 
fome convenient 
lay down the rhombus Q*':- , 
(BADC) fo that two cf \ 5 
its fides BA, BC are di- '\ 
reCted to the extremities 
of the line. Mark tlie 
interfeClions O and P, (as 
in the firft cafe of the pre¬ 
ceding problem;) then the B 
triangle ODP will be fimilar to the triangle RBQ. 
each of the redangles DOX'BQ, DPXBR, being 
to the fquare on the fide of the rhombus (as in'the 
preceding prob.) they muft therefore be equal to each 
other, or DOxBQ=DPxBR; therefore DO : DP :: 
BR : BQ ; and lince the angles at D and B are equal, 
the triangles ODP, RBQ will be fimilar. Therefore 
OD : OP :: RB : RQ. 
Suppofe ODpigf. 5in. PD=iif. loin. 0P=:i3f. 7in. 
and die fide of the rhombus =2 100 feet; 
. I0000 ^ 
then : 100 : : 100 :-—- = RB; th-erefore 
iifl ' 
9J-(0D) : i3^(0P) :: if22?(RB) ; 
1219 feet = RQ. 
Therefore- 
