triangle may be made: thus,-from the extremities of the base draw arcs with a common radius equal to the equal sides of the triangle, and join the points A, B, and C.

3. A figure approximating to an oval may be drawn by describing, from the extremities of a given line AB, equal circles intersecting in C and F, and by taking the diameter A E, D or BD, and from C and F drawing the arcs GH and IK to meet the arcs IG and KH; the figure DGHEKID will approximate to an oval.

A

D

4. By drawing an equilateral triangle on wood or brass, an instrument, BDE, or CFG, may be made, with which an inaccessible distance can be measured. Let A be an object on the other side of a river; at station B place the instrument so that A can be seen along the edge BD; then without changing the position of the instrument, look along the other edge BE, and set out a line of indefinite length as a continuation of BE; carry the instrument along the line BC, with its edge BE upon the B line BC, until by looking along the edge DE brought to coincide with CF, the same object A will appear in a straight line with CF: then A, B, and C will be the angular points of an equilateral triangle; and as the sides are equal, by measuring from the station B to the station C, the distance from B to A, or from C to A, will be ascertained.

E

G C

PROP, 2.-PROB.

From a given point to draw a straight line equal to a given straight

line.

SOLUTION.-Pst. 1. A st. line may be drawn from any point to any other point. P. 1. On a given line to desc. an equil. triangle.

Pst. 2. A terminated st. line may be produced to any length in a st. line.

Pst. 3. A circle may be drawn from any centre at any distance from that centre.

DEMONSTRATION. -Def. 15. All st. lines from the centre of the circle to the circumference, are equal.

Ax. 3. If equals be taken from equals, the remainders are equal.
Ax. 1. Things which are equal to the same thing, are equal each to
the others.

from centre B with BC desc. the OCHG; and from centre D with DG desc. O GLK;

then the st. line AL from A the given line
BC.

B is the centre of CGH, and D of O
GLK;

.. BG BC, and DL

But the line DA

DG.

DB;

and taking away these equals from the equals

DL and DG,

the rem. AL the rem. BG:

But BG being = BC, the rem. AL = BC. Wherefore from the given point A has been drawn a st. line AL the given st. line BC.

SCHOLIUM.-When the given point is out of the given line, or of the given line produced, this problem admits of eight cases, each of which is a solution of the problem; but if the given point is in the given line, or in the given line produced, of only four cases.

=

It is very conducive to the learner's improvement, when the proposition admits of it, to vary the mode of solution: of the eight cases mentioned, we will take another, in which the given point A is joined to C, the other extremity of the line BC.

The same method will be pursued in the SOLUTION: join A and C, and on AC construct an equil. triangle ADC; produce its sides to

E

K

H

Q.E.F.

C

B

F

E and F; and with CB as radius desc. the circle GBH, and with DG as radius, the circle GLK; the line AL will be drawn equal to CB.

The DEMONSTRATION follows the same course as in the first case given above. The learner may solve some of the other cases for himself.

USE AND APPLICATION.-Practically the given distance BC will be taken in the compasses, or measured by a string, or some instrument, as a foot, or a yard, and a line of the required length be marked off from A, as AL

PROP. 3-PROB.

From the greater of two given lines to cut off a part equal to

the less.

SOLUTION.-P. 2. From a given point to draw a line equal to a given line.

Pst. 3. A may be described from any centre at any distance from that centre.

DEMONSTRATION.-Def. 15. All straight lines from the centre to the circumference of a O, are equal.

Ax. 1. Magnitudes which are equal to the same magnitude, are equal to each other.

EXP. 1 Data. | Given the st. lines

AB and C, AB

being the greater

of the two;

2 Quæs. it is required to cut

off from AB a part

H

EB

the less C.

CONS. 1 P. 2.

From the point A

F

draw a st. line AD
= to C;

'G

2 Pst. 3. and from A with ra

dius AD describe the ODEF;
then the st. line AE cut off from AB,
less.

A is the centre of the O FED,
2 Def. 15. the st. line AE the st. line AD:

3 Sol.

DEM. 1 C. 2.

3 C. 1.

4 Ax. 1.

5 Rec.

But the st. line AD

the given st. line C, consequently, the st. line AE, cut off from AB, the st. line C.

Therefore from AB, the greater has been cut off,

&c.

Q.E.F.

SCHOLIUM.-A less line AD may be made equal to a greater AB, by describing a circle GBH with the radius AB, and producing AD until it meets the circle in H; then AH will equal A B.

USE AND APPLICATION.-1. This problem is performed practically, by transferring the distance C from A on AB.

2. The Problems 2 and 3 are of very frequent application, for in Geometry we are continually required to draw a line equal to a given line; or to take away from a greater line a part equal to the less; or to lengthen the less so that with the part produced it may equal the greater.

3. The 3rd Problem furnishes the means of constructing a Scale of Equal Parts; thus, Take AB of indefinite length towards B, and let C be the given st. line or part that is to be cut off from AB; from AB cut off a part equal to C, as AE; then again from EB another part equal to C, as EF; and so on; the parts in A B are each equal to C and to one another; and AB is a scale of equal parts; for the radii AD, ED', FD", GD", being each equal to C, AE, EF, FG, and GB are all equal.

On the same principle we take a line KL, and from one extremity K set off on the line ten equal parts, as in KO; then from O set along the line parts each equal to KO: if the parts between K and O are tenths, the parts 1, 2, 3, 4, 5 will be units; but if the parts between K and O are units, then the parts numbered 1, 2, 3, 4, 5 will be tens, namely, 10, 20, 30, 40, 50. By a scale of this kind, the comparative lengths of lines may be readily measured.

For the advantageous use of a Scale of Equal Parts, we should understand the nature of Representative Values. A miniature, of not more than a square inch in surface, is representative of the human face; and a map, on a square of 12 inches, may be representative of a tract in the heavens that takes in distances which we can scarcely conceive. The lines in the miniature and in the map, are in due proportion to those in the face and in the expanse of heaven; and thus they possess a representative value,-they are not the actual distances, but they stand for them. The inch on a scale may indicate a mile, or a thousand miles of distance; but if each portion of a mile, if each mile, or thousand miles, is given of the same relative size, then the map is a true representation; stretch out all its parts in an equal degree, and at last by superposition it would actually cover every point of the wide surface of which it is but the mark or outline.

By such a use of the Scale of Equal Parts, and of the Scale for Angular Magnitude, we can construct figures that are a true index of the positions and real distances of cities, mountains, and seas, and in some respects of the constellations of heaven. For instance, -if by actual observation and measurement it is ascertained that there are two towns each distant thirtyfive miles from a third, and that the two, in reference to the third, are at an angle of 21° apart, a plan may easily be drawn which shall correctly shew their situation with respect From L draw T

to each other.

a line L M of 35 from a scale of equal parts; at L with a semicircle make an angle of 21°; and along LT from the same scale set another 35 ;the points L, M, T will represent the situations and distances of the three towns.

M

210

PROP. 4.-THEOR.-Important.)

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to each other; they shall likewise have their bases, or third sides, equal; and the two triangles shall be equal, and their angles shall be equal, each to each, viz., those to which the equal sides are opposite.

DEMONSTRATION.-Ax. 10. Two straight lines cannot enclose a space. Ax. 8. Magnitudes which coincide with one another are equal. EXP. |1| Hyp. 1. In the As ABC,DEF,

2

=

let AB DE, and
AC=DF;

2. Also let the included
L BAC the inclu-
ded / EDF;

3 Concl. 1. then the base BC

=

D

AA

2. and▲ABC=ADEF; B

3. also ABC= DEF,

and ACB / DFE.

=

CE F

DEM. 1 Superp. For, applying ▲ ABC to ▲ DEF, so that A

is on D, and AB on DE;

2 D.1, H.1 AB coincides with and is equal to DE,

3 Ax. 8.. the B shall coincide with the E: