PRACTICAL APPLICATIONS.

PROBLEM I.

To bisect a given straight line.

Let AB be a given straight line. From A and B, as centres, with a radius greater than one half of AB, describe arcs intersecting at E and F: join E and F, by the straight line EF. Then will EF bisect the given line AB. For, E and F are each equally distant from A and B; and consequently, the line line EF bisects AB (B. I., P. XVI., C.).

PROBLEM II.

To erect a perpendicular to a given straight line, at a given point of that line.

Let EF be a given line, and let A be a given point of that line.

From A, lay off the equal distances AB and 'AC; from B and C, as centres, with a radius greater than one half

E

F

of BC, describe arcs intersecting at D; draw the line AD: then will AD be the perpendicular required. For, D and A are each equally distant from B and C; consequently, DA is perpendicular to BC at the given point A (B. I., P. XVI., C.).

PROBLEM III.

To draw a perpendicular to a given straight line, from a given point without that line.

A

Let BD be the given line, and A the given point. From A, as a centre, with a ra dius sufficiently great, describe an arc cutting BD in two points, B and D; with B and D as centres, and a radius greater than one-half of BD, describe arcs intersecting at E; draw AE then will AE be the perpendi

:

cular required. For, A and E are each equally distant from B and D: hence, AE is perpendicular to BD (B. I., P. XVI., C.).

At a point on a given straight line, to construct an angle equal to a given angle.

Let A be the given point, AB the given line, and

IKL the given angle.

From the vertex K as a centre, with any radius KI, describe the arc IL, terminating in the sides of the angle.

K

D

B

From A as a centre, with a radius AB, equal to KI,

describe the indefinite arc BO; then, with a radius equal

to the chord LI, from B as a centre, describe an arc cutting the arc BO in D;

draw AD: then will BAD be equal to the angle K.

For, the arcs BD, IL,

have equal radii and equal

K

B

chords hence, they are equal (P. IV.); therefore, the angles BAD, IKL, measured by them, are also equal (P. XV.).

PROBLEM V.

To bisect a given arc, or a given angle.

1o. Let AEB be a given arc, and C its centre.

Draw the chord AB; through C,

draw CD perpendicular to AB (Prob.

III.): then will CD

C

bisect the arc

explained : then will CD

bisect the angle ACB. For, the arcs AE and EB are equal, from what was just shown; consequently, the angles ACE and ECB are also equal (P. XV.).

Scholium. If each half of an arc or angle be bisected, the original arc or angle will be divided into four equal parts; and if each of these be bisected, the original arc or angle will be divided into eight equal parts; and so on.

PROBLEM VI.

Through a given point, to draw a straight line parallel to a given straight line.

Let A be a given point, and BC a given line.
From the point A as a centre,

with a radius AE, greater than the
shortest distance from A to BC,
describe an indefinite arc EO; from
E as a centre, with the same ra-
dius, describe the arc AF; lay off

B

ED equal to AF, and draw AD: then will AD be the parallel required.

For, drawing AE, the angles AEF, EAD, are equal (P. XV.); therefore, the lines AD, EF are parallel (B. I., P. XIX., C. 1.).

PROBLEM VII.

Given, two angles of a triangle, to construct the third

angle.

Let A and B be given angles of a triangle.

Draw a line DF, and at some

point of it, as E, construct the an

gle FEII equal to A, and HEC

equal to B. Then, will CED be D

equal to the required angle.

H

For, the sum of the three angles at E is equal to two right angles (B. I., P. I., C. 3), as is also the sum of the three angles of a triangle (B. I., P. XXV.). Consequently, the third angle CED must be equal to the third angle of the triangle.

PROBLEM VIII.

Given, two sides and the included angle of a triangle, to construct the triangle.

Let B and C denote the given sides, and A the given.

GH: then will DGH be the required triangle (B. I., P. V.).

PROBLEM IX.

Given, one side and two angles of a triangle, to construct the triangle.

The two angles may be either both adjacent to the given side, or one may be adjacent and the other opposite to it. In the latter case, construct the third angle by Problem VII. We shall then have two angles and their included side.

Draw a straight line, and on it

lay off DE equal to the given side; at D construct an angle equal to

one of the adjacent an

gles, and at E construct an angle

equal to the other adjacent angle;

D

produce the sides DF and EG till they intersect at H: then will DEH be the triangle required (B. I, P. VI.).