B

A

D

E

At station A with a theodolite ascertain the angle CAD,-suppose 45°; measure AB, 300 feet, and at station B measure the angle CBD, 30°; and from these data construct the figure required.

:

From B draw a line of indefinite length BE, and take on it 300 equal parts to A at B draw an angle CBD of 30°; and at A an angle CAD of 45° the lines BC and AC intersect in the point C; and a perpendicular from C to BE, namely CD, will represent the perpendicular height of the mountain. Take CD in the compasses, and apply the distance to the same scale; it will be found that CD measures 400, feet or yards, according to the unit of length in A B.

3. On the principle that all the interior angles of any rectilineal figure are equal to twice as many rt. angles as the figure has sides, diminished by four rt. angles for the amount of the angles at the centre, we are enabled to construct any regular right-lined figure. The angular magnitude of each figure is equal to two rt. angles multiplied by the number of sides, minus four rt. angles; and the remainder divided by the number of sides or angles, gives each single angle of the regular figure. Let S equal the number of sides; 180° the measure of two rt. angles; and 360° the measure of four rt. angles; and let A represent the angle at each pair of sides, and C the angle at the centre by radii to the extremities of each side.

(S × 180°)-360
S

G

F

E

To construct a regular Polygon :-Ascertain by the foregoing formula the angle C, and the angle A. 1st. When the side AB is not given--Draw any radius, as CA, and at C, with another radius of the same length as the first, make the ang. C as ascertained by the formula; then describe with CA or CB the circle, and the distance AB will divide the circumference into as many parts as there are units in S join the points, and the polygon is formed. 2nd. When the side AB is givenAt A and B make angles equal to ang. A, as ascertained by the formula; and bisect those angles by the lines AC and BC meeting in the point C; triangle CAB having the angles CAB and CBA equal, the sides CA and CB are also equal: if now, with CA or CB as radius, a circle be described, AB will cut off as many arcs from the circumference as there are units in S;-draw the chords to the respective arcs, and the polygon will be completed.

B

PROP. 33.-THEOR.

The st. lines which join the extremities of two equal and parallel st. lines towards the same parts, are also equal and parallel.

CONSTRUCTION.-Pst. 1. A line may be drawn from one point to another. DEMONSTRATION.-P. 29. A line falling on two parallel lines makes the alternate angles equal.

P. 4. Two triangles having in each two sides and their included angle respectively equal, are equal in every other respect.

P. 27. If a line falling on two st. lines make the alternate angles equal, the two lines are parallel.

LABC

BCD:

=

AB CD, BC is common, and
BCD,

=

:. LACB = [ CBD, and AC

=

BD.

And BC with AC, BD, makes / ACB =
LCBD,

. the st. line AC is || to the st. line BD;
And AC is equal to BD.

Wherefore, the st. lines which join the extremities, &c.

Q.E.D.

SCHOLIUM.-A Parallelogram is a four-sided figure of which the opposite sides are equal and parallel, and the diagonals join opposite angles.

distance BH: repeat the process as often as the case requires;-then DA + EB, &c., equals the horizontal distance CG; and DB + EC, &c., the altitude AG.

PROP. 34.-THEOR.

The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them; that is, divides them into two equal parts.

DEMONSTRATION.-P. 29. A line falling on two parallel lines makes the alternate angles equal.

P. 26. If two triangles have two angles and a side of one equal to two angles and a side of the other, the other sides and angle are respectively equal.

Ax. 2. If equals be added to equals, the wholes are equal.

P. 4. If two triangles have two sides and the included angle equal in each, they are altogether equal.

8 Concl.

9 H. & D. 1.

10 P. 4.
11 Recap.

LACB,

=

L BCD, and CBD

..the whole ABD = the whole / ACD; and / BAC = LBDC;

=

..the opp. sides and angles ofs are equal. Also, AB = CD, BC is common, and ABC BCD;

=

Wherefore, the opposite sides and angles of parallelograms, &c.

Q.E.D.

SCHOLIUM.-1. The two diagonals, AD and BC, bisect each other.

2. The converse to Prop. 34 is,—If the opposite sides or opposite angles of a quadrilateral figure be equal, the opposite sides shall be parallel; i. e., the figure shall be a parallelogram.

3. Both the diagonals, AD and BC, being drawn, it may, with a few exceptions which require subsequent propositions, be proved, that a quadrilateral figure which has any two of the following properties, will also have the others:

These ten data, being combined in pairs, will give 45 distinct pairs; with each of these pairs it may be required to establish any of the eight other properties, and thus 360 questions, respecting such quadrilaterals, may be raised. These questions will furnish the student a useful geometrical exercise. "The 9th and 10th data require the aid of snbsequent propositions." -LARDNER'S Euclid, p. 49.

USE AND APPLICATION.-1. The construction and accuracy of the parallel ruler depends on Prop. 34.

2. A finite st. line may be divided into any given number of equal parts. EXP. 1 Data.

2 Quæs.

CONS. 1 Pst. 1.

Given a line AL,
and the number
of parts, as four;
to divide AL into

those parts.

From one extremity

of AL, as A, draw
an indefinite line
AX,

EX

D

C

B

2 P. 3 & Pst. 1. take AB, and make BC, CD, DE each = AB, and

3 P. 31.

4 Sol.

5 P. 31.

join EL;

through D, C, B draw Dd, Cc, Bb, s to EL;

the line AL is so divided that Abbc, bc=cd, and cd=d L.

Also through b draw bm || AX.

DEM. 1 C. 3 & 5, P. 34 Bb Cm is a,:.bm = BC or AB;

2 P. 29.

3 P. 26.

4 Sim.

also A cbm, and AbB = bcm:
therefore Ab is equal to bc.

And in the same way dcbc, and dL = cd.

Q.E.D.

3. On the same principle the Sliding Scale, called from the inventors the Vernier or Nonius, is constructed. This scale is very useful, as in the

Barometer, for measuring the hundredth part of an inch; or in the Theodolite, for measuring the minutes into which a degree is divided.

10. For measuring the hundredth part of an inch, we may take, because of its clearness, Ritchie's description, in his Geometry, p. 32.

"Suppose an

inch AB, divided into 10 equal parts, each part will be of an inch. Again, suppose a line CD equal to 9 of these parts to be also divided into 10 equal parts, each of these will be of of an inch, which is 10. But the length of one of the first divisions being, or, and that of the second, one of the first divisions is of an inch longer than one of the second. If the line CD slide along parallel to AB till the two divisions marked 1, 1, form a continuous line, the sliding scale will have moved 10 part of an inch towards B. If it slide along till the next two divisions coincide, it will have moved of an inch, &c." 100

20. For measuring the minutes into which a degree on a circle is divided. "If 59 degrees on the circumference or limb of the instrument be divided into 60 equal parts, the difference between the length of one degree and one of the latter divisions will obviously be of a degree, or one minute. This kind of Vernier, on account of its great length, is seldom employed. If each degree on the arc AB of 15° be divided into two equal parts or half degrees, and if 29 of these be taken and divided into 30 equal parts, as in CD, the difference between the length of half a degree and one of the new divisions in CD will obviously be of half a degree, or

10

30

of a degree, that is, one minute. This is a very common vernier." If the limb CD move along the arc AB, so that the first division on CD and the first on AB are in one and the same line, the vernier will have moved of, or of a 14 degree, i. e., one minute; if the second division on CD coincide with the second on the arc AB, the vernier will have sol moved of a degree, or two minutes, &c.

4. When two objects, A and B, are inaccessible from each other, with an instrument to set out perpendiculars from them to a line CD, parallel to the line joining A and B, the distance may be obtained; for, by Prop. 34, CD is equal to AB; if therefore CD be measured, we learn the distance from A to B.

A

CA

10°

D

B

5. Also, when there intervenes an obstacle, as a house, or a lake, to the continuation of a st. line AB, beyond the obstacle H ;-the direction of A B may be ascertained by the method of taking perpendiculars, and of drawing parallel lines.