PROBLEMS

RELATING TO THE FIRST TWO BOOKS.

PROBLEM I.

To bisect a given straight line AB.

D

E

B

From the points A and B as centres, with a radius greater than the half of AB, describe two arcs cutting each other in D; the point D will be equally distant from A and B. Find, in like manner, above or beneath the line AB, a second point E, equally distant from the points A and B; through the two points D and E, draw the line DE: it will bisect the line AB in C.

Ж

E

For the two points D and E, being each equally distant from the extremities A and B, must both lie in the perpendicular raised from the middle of AB. (16. 1.) But only one straight line can pass through two given points; (Ax. 12;) hence the line DE must itself be that perpendicular, which cuts AB into two equal parts at the point C.

PROBLEM II.

At a given point A, in a given straight line BC, to erect a perpendicular to that line.

Take the points B and C at equal distances from A; then from the points B and C as centres, with a radius greater than BA, describe two arcs intersecting each other in D; draw AD: it will be the perpendicular B required.

A

For the point D, being equally distant from B and C, must lie in the perpendicular raised from the middle of BC; (16. 1;) therefore AD is that perpendicular.

Scholium. The same construction serves for making a right-angle BAD, at a given point A, on a given straight line BC.

PROBLEM III.

From a given point A, without the straight line BD, to let fall a perpendicular on that line.

From the point A as a centre, and with a radius sufficiently great, describe an arc cutting the line BD in the two points B and D; then mark a point E, equally distant from the points B and D, and draw AE: it will be the perpendicular required.

B

A

E

C

-D

For the two points A and E, are each equally distant from the points B and D; hence the line AE is a perpendicular passing through the middle of BD. (16. 1.)

PROBLEM IV.

At the point A in a given line AB, to make an angle equal to the given angle K.

From the vertex K as a centre, with any radius, describe the arc IL, terminating in the two sides of the angle; from the

44

I A

point A as a centre, with a distance AB equal to KI, describe the indefinite arc BO; then take a radius equal to the chord

LI, with which, from the point B as a centre, describe an arc cutting the indefinite one BO in D; draw AD; and the angle DAB will be equal to the given angle K.

For the two arcs BD, LI have equal radii, and equal chords; hence they are equal; (4.2;) consequently the angles BAD, IKL measured by them, are equal.

each equally distant from the extremities A and B of the chord AB; hence the line CD bisects the chord at rightangles; (16.1;) therefore it must bisect the arc AB in the point E. (6. 2.)

Second. Let it be required to bisect the angle ACB. From the vertex C as a centre, describe the arc AB; which is then bisected as above. It is plain that the line CD will divide the angle ACB into two equal parts. (11. 2.)

Scholium. By the same construction, each of the halves AE, EB may be divided into two equal parts; and thus by successive subdivisions, a given angle or a given arc may be divided into four equal parts, into eight, sixteen, and so on.

Through a given point A, to draw a parallel to a given straight

with the same radius, describe the arc AF; take EO=AF, and draw AD: this will be the parallel required.

For, joining AE, the alternate angles AEF, EAD are equal; (11.2;) therefore the lines AD, EF are parallel. (21. 1.)

PROBLEM VII.

Two angles A and B of a triangle being given, to find the third. Draw the indefinite line DEF;

D

C

H

E

F

at any point E, make the angle DEC=A, and the angle CEH B: the remaining angle HEF will be the third angle required; because those three angles are together equal to two right-angles, (1. 1. Cor. 3,) therefore they are equal to the three angles of a triangle. (28. 1.)

PROBLEM VIII.

Two sides B and C of a triangle, and the angle A which they

contain, being given, to describe the triangle.

Having drawn the indefinite line DE; at the point D, make the angle EDF equal to the given angle A; then take DG=B, DH-C, and draw GH; then DGH will be the triangle required. (5. 1.)

B C

E

G

I

H

PROBLEM IX.

A side and two angles of a triangle being given, to describe the triangle.

The two angles will either be both adjacent to the given side, or the one adjacent and the other opposite. In the

G

F

H

latter case, find the third angle; (Prob. 7.2;) and the two adjacent angles will D

E

thus be known. In the former case, draw the straight line DE equal to the given side; at the point D, make an angle EDF equal to one of the adjacent angles, and at E, an angle DEG equal to the other; the two lines DF, EG, will cut each other in H; then will DEH be the triangle required. (6. 1.)

PROBLEM X.

The three sides A, B, C of a triangle being given, to describe the triangle.

Draw DE equal to the side A; from the point E as a centre, with a radius equal to the second side B, describe an arc; from D

as a centre, with a radius equal to the third side C, describe another arc intersecting the former in F; draw DF, EF; then will DEF be the triangle required. (10. 1.)

Scholium. If one of the given sides A were greater than the sum of the other two, B and HC, constructing the figure as above, the arcs BF, CH, would not intersect each other; hence

D

E

Br

CH

F

D

-E

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