GEOMETRY. 
8-13 
EF, and the intersection 0 will be the centre ] 
required. 
Trob. 10. To draw a tangent to a given 
circle that shall pass through a given point, 
A. 
From the centre O, draw the radius OA. 
Through the point A, draw DE perpendicu- 
lar to OA, and it will be the tangent re- 
quired. 
Prob. 11. To draw a tangent to a circle, or 
any segment of a circle ABC, through a given 
point B, without making use of the centre of 
the circle. 
l ake any two equal divisions upon the cir- 
cle from the given point B, towards d and e, 
and draw the chord eB. Upon B, as a centre, 
■ with the distance Bd, describe the arc fdg, 
cutting the chord eB in f. Make dg equal 
to df; through g draw gB, and it will be the 
tangent required, 
Prob. 12. Given three points, A, B, C, not 
in a straight line, to describe a circle that 
j shall pass through them. 
Bisect the lines AB, BC, by the perpendi- 
■ culars ad, bd, meeting at d. Upon d, with 
' the distance d A, d B, describe ABC, and it 
j will be the required circle. 
Prob. 13. To describe the segment of a 
circle to any length AB, and height CD. 
Bisect AB by the perpendicular Dg, cut- 
j ting AB in c. From c make cD on the per- 
j pendicular equal to CD. Draw AD, and 
bisect it by a perpendicular ef, cutting Dg 
j in g. Upon g the centre, describe ADB, 
and it will be the required segment. 
Prob. 1 4. In any given triangle to inscribe 
| a circle. 
Bisect any two angles, A and C, with the 
lines AD and DC. From D the point of in- 
I tersection, let fall the perpendicular DE; it 
I will be the radius of the circle required. 
Prob. 13. In a given square, to describe a 
regular octagon. 
Draw the diagonals AC and BD, intersect- 
ing at e. Upon the points A, B, C, 1), as 
centres, with a radius eC, describe the arcs 
ken, meg, fei, &c. Join fn, mb, ki, 
lg, and it v ill be the required octagon. 
Prob. 16. In a given circle, to describe 
any regular polygon. 
Divide the circumference into as many 
parts as there are sides in the polygon to be 
drawn, and join the points of division. 
Prob. 17. Upon a given line AB, to con- 
struct an equilateral triangle. 
Upon the points A, and B, with a radius 
| equal to AB, describe arches cutting each 
other at C. Draw AC and BC, and ABC 
wiil be the triangle required. 
Prob. 18. To make a trapezium eqital and 
similar to a given trapezium ABCI). 
Divide the given trapezium ABCD into 
two triangles by the diagonal DB. Make 
EF equal to AB ; upon EF construct the 
triangle EFTI, whose sides shall be re- 
spectively equal to those of the triangle 
ABD by the last problem. Upon IJF, 
which is equal to DB, construct the triangle 
HFG, whose sides are respectively equal to 
DBG ; then EFG.H wall be the trapezium 
required, 
By the help of this problem any plan may 
be copied ; as every figure, however irregular, 
may be divided into triangles. Upon this the 
practice of land-surveying and making plans 
of estates, is founded. 
Prob. 19. To make a square equal to two 
given squares. Make the sides DE and DF 
of the two given squares A and B, form the 
sides of a right-angled triangle FDE ; draw 
the hypothenuse FE ; on it describe the 
square EFGII, and it will be the square re- 
quired. 
Prob. 20. Between two given lines, AB 
and CD, to find a mean proportional. 
Draw the right line, EG, in which make 
EF ecp.ial to AB, and FG equal to CD. Bi- 
sect EG in H, and with H E or HG, as ra- 
dius, describe the semicircle E1G. From F 
draw FI perpendicular to EG, cutting the 
circle in I ; and IF will be the mean propor- 
tional required. 
Geometry, application of algebra to. 
When a geometrical problem is proposed to 
be resolved by algebra, you are, in the first 
place, to describe a figure that shall repre- 
sent, or exhibit, the several parts or condi- 
tions thereof, and look upon that figure as the 
true one ; then, having considered attentively 
the nature of the problem, you are next to 
prepare the figure for a solution (if need be), 
by producing and drawing such lines therein 
as appear most conducive to that end. This 
done, let the' unknown line or lines which 
you think will be the easiest found (whether 
required or not), together w'ith the known 
ones (or as many of them as are requisite), be 
denoted by proper symbols ; then proceed to 
the operation, by observing the relation that 
the several parts of the, figure have to each 
other ; in order to which a competent know- 
ledge of the elements of geometry is abso- 
lutely necessary. 
As no general rule can be given for the 
drawing of lines, and electing the most pro- 
per quantities to substitute for, so as to always 
bring out the most simple conclusions (be- 
cause different problems require different me- 
thods of solution) ; the best way, therefore, 
to gain experience in this matter, is to at- 
tempt the solution of the same problem se- 
veral ways, and then apply that which suc- 
ceeds best to other cases of the same kind 
when they afterwards occur. We shall, how- 
ever, subjoin a few general directions, which 
will be found of use. . 
1. In preparing the figure, by drawing 
lines, let them be either parallel or perpendi- 
cular to other lines in the figure, or so as to 
form similar triangles ; and, if an angle be 
given, let the perpendicular be opposite to 
that angle, and also fall from the end of a 
given line, if possible. 
2. In electing proper quantities to substi- 
tute for, let those be chosen (whether re- 
quired or not) which lie nearest the known, 
or given parts of the figure, and by help 
whereof the next adjacent parts may be ex- 
pressed, without the intervention of surds, 
by addition and subtraction only. Thus, if 
the problem were to find the perpendicular of 
a plane triangle, from the three sides given, 
it will be much better to substitute for one of 
the segments of the base than for the perpen- 
dicular, though the quantity required ; be- 
cause the whole base being given, the other 
segment will be given, or expressed, by sub- 
traction only, and so the final equation comes 
out a single one ; from whence the segments 
being known, the perpendicular is easily 
found by common arithmetic: whereas, if 
the perpendicular were to be first sought, 
both the segments would be surd quantities, 
5 O 2 
and the final equation an unsightly qua- 
dratic one. 
3. When, in any problem, there are two 
lines <or quantities alike related to other parts 
of the figure, or problem, the best way is to 
make use of neither of them, but to substi- 
tute for their sum, their rectangle, or the 
sum of their alternate quotients, or for some 
line or lines in the figure, to which they have 
both the same relation. 
4. If the area, or the perimeter, of a figure 
be given, or such parts thereof as have but a 
remote relation to the parts required, it wiil 
sometimes be of use to assume another 
figure similar to the proposed one, whereof 
one side is unity, or some other known quan- 
tity ; from whence the other parts of this 
figure, by the known proportions of the ho- 
mologous sides, or parts, may be found, and 
an equation obtained. 
These are the most general observations, 
which we shall now proceed to illustrate by 
examples. 
Prob. I. The base (/>), and the sum of the hypothenuse 
and perpendicular (<z) of a right-angled trianvlc 
ABG> being given] tofnd the perpendicular. See 
Fig. 36. 
Let the perpendicular BC be denoted by *•; 
then the hypothenuse AC will be expressed by 
a — .v ; but (by Euc. 47. 1.) AB 2 -j- BC 2 = AC 7 ; 
that is, b 2 -|- x 2 — a 2 — 2 ax -j- xx ; whence a: 
— — the perpendicular required. 
Prob. II. The diagonal, and the perimeter of a rect- 
angle, ABCD, being given ; to find the sides. See 
Fig. 37. 
Put the diagonal BD — a, half the perimeter 
(DA -j- AB) zz b, and AB =z x] then will AD 
— b — x ; and therefore, AB 2 -j- AD 2 being — ; 
BD 2 , we have x 2 -j- b 2 — 2bx -\- x 2 =. b l \ which, 
solved, gives x — 
e/2 a 2 — b 2 Jf-b 
2 ‘ 
Pro*. III. The area of a right-angled triangle ABG, 
and the sides of a rectangle EBDF, inscribed therein , 
being given ; to determine the sides of the triangle. 
See Fig. 38. 
Put DF = a, DE z= b, BC =r x, and the meal 
sure of the given area ABC = d ; then, by simi- 
lar triangles, we shall have x — b (CF) * a (DF) 
** x (BC) , AB = Therefore 
x — b a — b 
X — = d, and consequently ax 2 — 2 dx — 2bd } 
, 5M* 2 bd . 
or ar zz. ; which, solved, gives x 
n n ° 
2bd 
a 
from whence AB and 
AC will likewise be known. 
Prob. IV. Having the lengths of the three perpendi- 
culars, PF, PG, PH, dranvn from a certain point 
P, •within an equilateral triangle ABC, to the three 
sides thereof : from thence to determine the sides. See 
Big. 39. 
Let lines be drawn from P to the three angles 
of the triangle ; and let CD be perpendicular to 
AB : call PF, a ; PG, b ; PH, c ; and AD = ar : 
then will AC (— AB) zz 2x, and CD (zz 
^/AC 2 — AD 2 ) = // 3xx zz: ar //3 ; and conse- 
quently the area of the whole triangle ABC 
(— CD X AD) ~ xx\f3. But this friangle is 
composed of the three triangles APB, BPC, and 
APC ; whereof the respective areas are ax, bx , 
and ex. Therefore we have xx 3 ezt ax -f- bx 
