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;

D

have equal radii and equal

:

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 bisect the arc AEB (P. VI.).

2o. Let ACB be a given angle. With C as a centre, and any radius CB, describe the arc bisect it by the line CD, as just

ᏴᎪ ;

explained then will CD bisect the angle ACB.

D

B

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 line parallel to a given

line.

point, and BC a given line.
as a centre,

B

F

1

E

Let A be a given From the point A with a radius AE, greater than the shortest distance from A to to BC, describe an indefinite arc EO; from E as a centre, with the same radius, describe the arc AF; lay off ED equal to AF, and draw AD: then will AD be the parallel required.

A

D

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

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 angle FEH equal to A, and HEC equal to B. Then, will CED be equal to the required angle.

H

D

E

F

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 angle.

Draw the indefinite line DF, and at D construct an angle FDE, equal to the angle A; on DF, lay off DH equal to the side C, and on DE, lay off DG equal to the side B; draw

G---E

A

D

H.-F

B

C

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

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

PROBLEM X.

Given, the three sides of a triangle, to construct the triangle.

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

Draw DE, and make it equal

D

AH

B

E

to the side A; from D as a centre, with a radius equal to the side B, describe an arc; from E as a centre, with a radius equal to the side C, describe an arc intersecting the former at F; draw DF and EF: then will DEF be the triangle required (B. I., P. X.).

CH

Scholium. In order that the construction may be possible, any one of the given sides must be less than the sum of the other two, and greater than their difference (B. I., P. VII., S.).

PROBLEM XI.

Given, two sides of a triangle, and the angle opposite one of them, to construct the triangle.

Let A and B be the given sides, and the given angle.

Draw an indefinite line DG, and at some point of it, as D, construct an angle GDE equal to the given angle; on one side of this angle lay off the distance DE equal to the side B adjacent to the given angle; from E as

A

Br

E

D

1

a centre, with a radius equal to the side opposite the given angle, describe an arc cutting the side DG at G; draw EG. Then will DEG be the required triangle.

For, the sides DE and EG are equal to the given sides, and the angle D, opposite one of them, is equal to the given angle.

Scholium. When the side opposite the given angle is greater than the other given side, there will be but one solution. When the given angle is acute, and the side opposite the given angle is less than the other given side, and greater than the shortest distance from E to DG, there will be two solutions, DEG and DEF. When the side opposite the given angle is

C

BH

E

D

F

G

equal to the shortest distance from E to DG, the arc will be tangent to DE, the angle opposite DE will be a right angle, and there will be but one solution. When the side opposite the given angle is shorter than the distance from E to DG, there will be no solution.

PROBLEM XII.

Given, two adjacent sides of a parallelogram and their included angle, to construct the parallelogram.

Let A and B be the given sides, and C the given angle.

allel to DF : then will DFGE be the parallelogram re

quired.