-GEOMETRY. 
For D being equally di from B and C, it must 
be in a line perpendicilarto middle of BC. There. 
fore AD is the perpendicular mreduty. cial YP 
By) 
i - ; 0) eda alana » Bae nD 
allt? ; . 64 Sopp yo! et oe Poel 
To draw a perpendicular toa given line BD from a 
given point A without that line. ~ 
On A asa centre, with a sufficiently great radius, 
describe an arc’to cut the line in Band D. Find next 
a point E.equally distant from A and B; join AE, meet- 
“ing BD in 'C, and AC will be the perpendicular re- 
“through D, and AD will 
mkt © > oskor 
quired 
anal For the two points A, E are each equally distant 
from B and D ; therefore AE is perpendicular to BD. 
4 
3 an 
“ > \i po Pros. VE 
SP Ata given point A in’a, 
angle equal to a given angle 
‘On K asa centre, with ai aoe , destribe an are 
“IL, to terminate in the’ dae of the angle. On A 
as a centre, with the same radius, describe an indefi- 
nite arc BO. On Bas a centre with a’ radius equal.to 
the chord of the are LI, describe .an arc to cut the. are 
BO in D; draw AD, and the angle DAB’ shall be 
* equal to the Ses arigle K. cages Ot wre 
“For IL, BD are equal chords in equal circles, there- 
fore the arc IL = are BD (4.), and the angle K = 
angle A, (15.) 
Wa 
gen Tine AB to make an 
eames Mies Od ently, lrg: 
To divide a given angle, or an are, into two equal 
(2. If it is'an are AB whose éentre is ‘C, which is to 
“be divided, on A and Bas centres, with the same ra- 
dius describe ares to meet in D’; th Cand D draw 
a esight line CD, which will bisect the are AB at the 
point EF. 
© °¥For C and D are each at the same distance from the 
extremities of the chord AB; therefore CD is h> 
dicular at the middle of the chord (17. 1.), and conse- 
- quently must bisect the arc AEB, ( 1. 6.) f 
2. Ifthe angle ACB is to be bisected; in the first place, 
an arc AEB is to be described on € the vertex’ of the 
‘angle as a centre ; then a point D must be found equal- 
tine A and B as before, and a line drawn 
rom C through D will bisect the angle. 
“For if the chord AB be drawn, CD will be perpen- 
dicular at the middle of AB ; therefore CD bisects the 
os ” (6.), and consequently bisects the angle ACB, 
¢ ie . t ; 4 
’ 
Pros, VIII. 
Through a given point A, to draw a straight line 
parallel to a given line BC, 
‘.. On A asa centre, with a radius sufficiently great, 
describe an indefinite arc EO; on Easa ett with 
the same radius, describe an arc AF ; also on E asa 
“centre with the chord of the arc AF as a radius, de- 
ibe an are to meet EO in D; draw a line from A 
be the parallel required. 
“VOL, X. PARTI. 
For AF and DE are manifestly equal arcs of equal 
circles ; therefore the angles BEA, DAE are equal ; 
and hence BC is parallel to AD, (Cor. 1. 20. 1.) 
Prop. IX. 
To describe a triangle, the sides of which shall be 
equal to three given straight lines A, B, C. 
Draw a straight line DE equal to one of the lines A; 
on E as a centre, with a radius equal to another of the 
lines :B,, describe an arc ; on Das a centre, with a ra- 
dius equal to'thes remaining line C; describe another 
are, cutting the former in F ; draw DF, EF; and DEF 
shall be the triangle required, as is sufficiently evident. 
Scrorrum: The problem is only possible when the 
suntof any two of the given lines is greater than the 
third, (8, 1) 
alanis’ entietasl 
(i Pros. Xs) > 
3 ww P53. : 
‘To construct a parallelogram, the adjacent sides of 
which may»be equal to two given lines A, B, and the 
angle which they contain equal toa given angle C. . 
© Draw'a straight line DEA ; at the point D) make 
anatgle FDE'= C, and) take DF = B; deseribe:two 
»ares, onevon: Fas ‘a. centre, with DE or A as a radius, 
and the other on E asa centre with B as aradius. From 
the point G, ‘where these ares cut each other, draw GF, 
GE, and DEGF will be the parallelogram required. 
For by the construction, the opposite sides are equal ; 
therefore the figure is a parallelogrant, (27. 1.) 
Con. If the given angle were aright angle, the figure 
would be a rectangle ;' and if the sides were equal, the 
figure would be a square. — 
Prop. XT. 
To find the centre of a circle, or of a given arc. 
Take any: three points A, B, C in the circumference 
of the circle, or in the are; join them by the lines AB, 
BC, (or suppose these lines drawn), and -bisect AB; BC 
by the, perpendiculars DE, FG (Prob. 8.) the point 
-O where these lines meet each other shall be the centre 
sought, as is evident from Prop. 7. 
Scnoirum. By this construction, a circle may be de- 
scribed through three given points ; or may be describ- 
ed about a given triangle ABC. ’ 
“Pros, XII. 
Through a given point A, to draw a tangent toa 
given circle. 
If the given point A is in the circumference (Fig.81.), 
draw the radius CA, and draw AD perpendicular to 
CA ; and DA shall be the tangent required, (9, 2.) 
Ifthe point A is without the circle, (Fig. 82.), draw 
AC to ie centre, bisect AC in O; and on O asa 
centre, with OC. as a radius, describe a circle which 
may cut the given circle in B; draw AB, and AB shall 
be the tangent required 
For join CB, and the angle ABC in a semicircle will 
be a right angle (19.) ; therefore AB, a perpendicular 
to the radius at its extremity B, is a tangent to the 
circle, (9.) fp 
fr 2D 
209 
Problems. 
—o 
Fig. 73. 
Fig. 79. 
Fig. 80. 
Figs.81,82- 
Fig. $1. 
Fig. 82. 
