In any plane triangle ABC, if the line CD be drawn perpendicular to the base AB, dividing it into two segments, AD, DB, and the base AB be bisected in the point H, we shall have,

As the base AB is to the sum of the sides, AC, BC, so is the difference of the sides to twice the distance DH of the perpendicular from the middle of the base.

Dem. With the greater side CB as radius, describe about the centre C the circle BFGE, meeting the other side produced in the points E and F, and the base AB produced in G; join GF and BE. Then AE is the sum, and AF the difference of the sides AC, CB; and since CD is perpendicular to GB, the line GB is bisected in D (by art. 43) and as AB is bisected in H, the line AG is equal to twice DH. Now in the triangles BAE, GAF, the angles ABE, GFA are equal (by art. 41) and the angle BAE is equal to GAF (by art. 28) therefore the remaining angles AEB, AGF, are equal, and the triangles BAE, GAF, are similar; consequently (by art. 54) AB: AE:: AF: AG, or twice HD, which is the proposition to be demonstrated. Having thus obtained HD, we may find the segments AD, DB, by adding HD to the half base HA or HB and by taking their difference.

LXI.

In any plane triangle, the square of radius is to the square of the co-sine of half of either of the angles, as the rectangle contained by the two sides including that angle. is to the rectangle contained by the A

half sum of the sides, and that half sum decreased by the side opposite to that angle.

Thus in the triangle CBE, the square of radius is to the square of the co-sine CB+CE+BE CB+CE-BE

of half the angle C, as the rectangle CBXCE is

to

2

2

For continue EC to A, making CA=CB, draw BD perpendicular to CE, bisect CE in H, and join AB. Then (supposing CB to be greater than EB) we CB3-BE2

have (by art. 60) CE: CB+BE : : CB-BE:

CE

=2.HD; by adding

half this to half the base-CH, we have the segment CD=

CB3-BE2+CE2

2.CE CB2-BE2+CE2+3 CE CB

to this adding CA or CB, we have AD=

2.CE.

CB+CE 3-BE3 CB+CE+BE>CB+CE-BE

Again, AD=AC+CD=CB+CD

2.CE

2.CE

hence AD 2=CB3+3CB·CD+CD2; also, BD2=CB2-CD2; hence AB2= AD3+BD3=2CB2+2CB.CD=2CBXCB+CD=3CB. AD; hence AB2: AD2 : :

CB: AD=

CB+CE+BE. CB+CE-BE

2.CE

but AB being radius, AD is the co-sine

of the angle A, which is equal to half the angle C (by art. 40 ;) therefore the square of radius is to square of the co-sine of half the angle C, as the rectangle CE.CB is CB+CE+BE CB+CE-BE

The other cases of this proposition may be demonstrated in the same

manner.

*

GEOMETRICAL PROBLEMS.

PROBLEM I.

To draw a Right Line CD parallel to a given Right Line AB, at any given distance, as at the point D.

C

D

B

WITH a pair of compasses take the nearest distance between the point D and the given right line AB; with that distance set one foot of the compasses any where on the line AB, as at A, and draw the arch C on the same side of the line AB as the point D, from the point D draw a line so as just to touch the arch C, and it is done; for the line CD will be parallel to the line AB, and at the distance of the point given D, as was required.

From a given point as C, to let fall a perpendicular
CO, on a given Right Line AB.

A

Take any extent in your compasses greater than the least distance between C and the given line AB; with one foot in C, describe an arch to cut the given A line AB in F and G;-with one foot in G describe an arch, and with the same distance, and one foot in F, describe another arch cutting the former in D; from C to D draw the line COD, cutting AB in 0; then CO will be the perpendicular required. E

PROBLEM V.

From a given point C to let fall a perpendicular CB on a given Line AB, when the perpendicular is to fall so near the end of the given line that it cannot be done as above.

Upon any point A of the line AB as a centre, and with the distance AC describe an arch E; choose any other point in the line AB, as D, and with the distance DC describe another arch intersecting the former in E, join CE cutting AB in B, and it is done, for CB will be the perpendicular required.

D

PROBLEM VI. To make an angle that shall contain any proposed number of degrees, from a given point in a given line. H CASE 1. When the given angle is right, or contains 900 let CA be the given line, and C the given point.

On C erect a perpendicular CD, and it is done;, for the angle DCA is an angle of 900. Or thus, on the point C as a centre, with the chord of 600* describe an arch GH, and set off thereon from G to H the distance of the chord of 90° and from C through H draw CHD, which will form the angle DCA of C 90° required.

CASE 2. When the angle is acute, as for example 36° 30' let CB be the given line and C the point at which the angle is to be made.

With the chord of 600 in your compasses, and one foot on C, as a centre, draw the arch FB, on which set off from B to F, the given angle 3640 taken from the line of chords; through F and the centre C draw the right line AC, and it is done; for the angle ACB will be an angle of 36° 30' as was required.

006

CASE 3. When the given angle is obtuse, as for example 127° 20' let CB be the given line and C the angular point.

H

E

G

Take the chord of 600 in your compasses, and with one foot on C as a centre, describe an arch BGHE, upon which set off the chord of 600 (which you already have in. your compasses) from B to G, and from G to H; then set off from G to E, the excess of the given angle above 600, which is 6730 taken from the line of chords; or you may set off from H to E, the excess of the given angle above 1200, which is 710; draw the line CE, and it is done, for the angle ECB will be an angle of 1270 20'.

127.20'

B

Were it required to measure a given angle, the process would have been nearly the same, by sweeping an arch as BE, and measuring it on the line of chords, as is evident.

PROBLEM VII.

To bisect a given arch of a circle AB, whose centre is C.

Take in your compasses any extent greater than the half of AB, and with one foot in A, describe an arch; with the same extent and one foot in B, describe another arch cutting the former in D; join CD and it is done, for this line will bisect the arch AB in the point E. It is also evident that the line CD bisects the angle C BCA, or divides it into two equal parts.

*For a description of the line of Chords see page 20.

D

E

A

PROBLEM VIII.

To find the centre of a given Circle. With any radius, and one foot in the circumference as at A, describe an arch of a circle, as CBD, cutting the given circle in B; with the same extent, and one foot in B, describe another arch CAD, cutting the former in C and D; through C and D draw the line CD, which will pass through the centre of the circle; in like manner, may another right line be drawn, as EFG, which shall cross the first right line at the centre required. This construction depends upon article 43 of Geometry.

PROBLEM IX.

To draw a Circle through any three given points not situated in a Right Line.
Let A, B and D be the given points. Take in
your compasses any distance greater than half

AB, and with one foot in A describe an arch E
EF; with the same extent, and one foot in B,
describe another arch cutting the former in the
points E, F, through which draw the indefinite
right line EFC; then take in your compasses
any extent greater than half BD, and with one
foot in B, describe an arch GH; with the same
extent, and one foot in D, describe another
arch cutting the former in the points G, H,
through which draw the right line GHC, cut-
ting the former right line EFC, in the point C;
upon the point C as a centre, with an extent
equal to CA, CB, or CD, as radius, describe
the sought circle.

PROBLEM X.

To divide a Circle into 2, 4, 8, 16, or 32, equal parts.

Draw a diameter through the centre, dividing the circle into two equal parts; bisect this diameter by another drawn perpendicular thereto, and the circle will be divided, into four equal parts or quadrants; bisect each of these quadrants again by right lines drawn through the centre, and the circle will be divided into eight equal parts; and so you may continue the bisections any number of times. This problem is useful in constructing the mariner's compass.

PROBLEM XI..

B

To divide a given Line into any number of equal parts.

Let it be required to divide the line AB into five equal parts. From the point A draw any line AD, making an angle with the line AB; then through the point B draw a line BC parallel to AD; and from A, withany small opening in your compasses, set off a number of equal parts on the line AD, less by one than the proposed number (which C number of equal parts in this example is 4:) 4 then from B set off the same number of the

2

3

H

D

same parts on the line BC, then join 4 and 1, 3 and 2, 2 and 3, 1 and 4, and these lines will cut the given line as required.

CONSTRUCTION OF THE PLANE SCALE.

1st. WITH the radius you intend for your scale, describe a semicircle ADB, (Plate II. fig. 1.) and from the centre C draw CD perpendicular to AB, which will divide the semicircle into two quadrants, AD, BD; continue CD towards S, draw BT perpendicular to CB, and join BD and AD.

2dly. Divide the quadrant BD into 9 equal parts, then will each of these be 10 degrees; subdivide each of these parts into single degrees, and if your radius will admit of it, into minutes or some aliquot parts of a degree greater than minutes.

Sdly. Set one foot of the compasses in B and transfer each of the divisions of the quadrant BD to the right line BD, then will BD be a line of chords. 4thly. From the points 10, 20, 30, &c. in the quadrant BD draw right lines parallel to CD, to cut the radius CB, and they will divide that line into a line of sines which must be numbered from C towards B.

5thly. If the same line of sines be numbered from B towards C, it will become a line of versed sines, which may be continued to 180°, if the same divisions be transferred on the same line on the other side of the centre C.

6thly. From the centre C, through the several divisions of the quadrant BD, draw right lines till they cut the tangent BT, so will the line BT become a line of tangents.

7thly. Setting one foot of the compasses in C, extend the other to the several divisions 10, 20, 30, &c. in the tangent line BT, and transfer these extents severally to the right line CS, then will that line be a line of secants.

8thly. Right lines drawn from A to the several divisions 10, 20, 30, &c. in the quadrant BD, will divide the radius CD into a line of semi-tangents.

9thly. Divide the quadrant AD into eight equal parts, and from A as a centre transfer these divisions severally into the line AD, then will AD be a line of Rhumbs, each division answering to 11°15' upon the line of chords. The use of this line is for protracting and measuring angles, according to the common division of the mariner's compass. If the radius AC be divided into 100 or 1000, &c. equal parts, and the lengths of the several sines, tangents, and secants, corresponding to the several arches of the quadrant, be measured thereby, and these numbers be set down in a table,* each in its proper coJumn, you will by these means have a collection of numbers by which the several cases in trigonometry may be solved. Right lines graduated as above, being placed severally upon a ruler, form the instrument called the Plane Scale, (see Plate II. fig. 2.) by which the lines and angles of all triangles may be measured. All right lines, as the sides of plain triangles, &c. when they are considered simply as such without having any relation to a circle, are measured by scales of equal parts, one of which is subdivided equally into 10, and this serves as a common division to all the rest. In most scales an inch is taken for a common measure, and what an inch is divided into is generally set at the end of the scale. By any common scale of equal parts, divided in this manner, any number less than 100 may be readily taken; but if the number should consist of three places of figures, the value of the third figure cannot be exactly ascertained, and in this case it is better to use a diagonal scale, by which any number consisting of three places of figures, may be exactly found. The figure of this scale is given in Plate II. fig. 3: its construction is as follows. Having prepared a ruler of convenient breadth for your scale, draw near the edges thereof two right lines, af, cg, parallel to each other; divide one of these lines as af, into equal parts, according to the size of your scale;† and

* In table XXIV. is given the sine and co-sine to every minute of the quadrant, to five places of de

cimals.

The length of one of these equal parts at the end of the scale to which this description refers is ab or cd. The length of one of the equal parts of the scale of the other end being the half of ab.