But the angle B DE (Theo. X.); hence the

BDE be equal to the angle B A C. is measured by half of the arc BE equal angle BAC is also measured by half the arc BE; that is, by half the difference of the arcs B E F and E F, or, since E F is equal to D F, by half the difference of the intercepted arcs BEF and D F.

PROBLEMS IN CONSTRUCTION.

169. The Problems in Construction, which follow in this book, require for their solution the application of the principles of the preceding books.

PROBLEM I. To bisect a given straight line, or to divide it into two equal parts.

Let A B be a straight line, which it is required to bisect.

C

E

A

-B

D

From the point A as a center, with a radius greater than the half of A B, describe an arc of a circle; and from the point B as a center, with the same radius, describe another arc, cutting the former in the points C and D. Through C and D draw the straight line CD; it will bisect AB in the point E.

For the two points C and D, being each equally distant from the extremities A and B, must both lie in the perpendicular raised from the middle point of A B (Theo. X. Cor. 2, Bk. I.). Therefore the line CD must divide the line A B into two equal parts at the point E.

PROBLEM II. From a given point, without a straight line, to draw a perpendicular to that line.

Let AB be the straight line, and let C be a given point without the line.

From the point C as a center, and with a radius sufficiently great, describe an arc cutting the line A B in two points, A and B; then, from the points A and B as centers, with a

radius greater than half of A B, describe

two arcs cutting each other in D, and draw the straight line CD; it will be the perpendicular required.

For, the two points C and D are each equally distant from the points A and B ; hence the line CD is a perpendicular passing through the middle of AB (Theo. X. Cor. 2, Bk. I.).

PROBLEM III. At a given point in a straight line to erect a perpendicular to that line.

Let A B be the straight line, and let D be a given point in it.

C

D

B

In the straight line A B, take the points A and B at equal distances from D; then from the points A and B as centers, with a radius greater than AD, describe two arcs cutting each other at C; through C and D draw the straight line CD; it will be the perpendicular required.

For the point C, being equally distant from A and B, must be in a line perpendicular to the middle of A B (Theo. X. Cor. 2, Bk. I.); hence CD has been drawn perpendicular to A B at the point D.

Scholium. The same construction serves for making a right angle, A D C, at a given point, D, on a given straight line,

ting the line A B at the points A and B ; through the point A, and the center D, draw the diameter A C. Then through C, where the diameter meets the arc, draw the straight line C B, and it will be the perpendicular required.

For the angle ABC, being inscribed in a semicircle, is a right angle (Theo. X. Cor. 2, Bk. III.).

PROBLEM V. At a point in a given straight line to make an angle equal to a given angle

center, with any radius, describe an arc, G E, terminating in the sides of the angle; from the point A as a center, with the same radius, describe the indefinite arc BD. Draw the chord GE; then from B as a center, with a radius describe an arc cutting the arc BD in C. the angle CAB will be equal to the given angle E F G.

equal to G E, Draw A C, and

For the two arcs, BC and G E, have equal radii and equal chords; therefore they are equal (Theo. II. Bk. III.); hence the angles CA B, EF G, measured by these arcs, are also equal (Theo. IV. Bk. III.).

PROBLEM VI. To bisect a given arc, or a given angle.

First. Let ADB be the given arc

which it is required to bisect.

Draw the chord AB; from the center C draw the line CD perpendicular to AB (Prob. III.); it will bisect the arc AD B in the point D.

For CD being a radius perpendicular

C

A

B

D

to a chord A B, must bisect the arc A D B which is subtended by that chord (Theo. V. Bk. III.).

Secondly. Let ACB be the angle which it is required to

bisect. From C as a center, with any radius, describe the arc ADB; bisect this arc, as in the first case, by drawing the line CD; and this line will also bisect the angle A C B.

For the angles ACD, BCD are equal, being measured by the equal arcs A D, D B (Theo. IV. Bk. III.).

Scholium. By the same construction, we may bisect each of the halves AD, DB; and thus, by successive subdivisions, a given angle or a given are may be divided into four equal parts, into eight, into sixteen, &c.

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

Let A be the given point, and

A

B

D

E

Then draw A B, making the angle E A B equal to the angle AEC (Prob. V.); and A B is parallel to CD.

For the alternate angles E AB, AEC, made by the straight line AE meeting the two straight lines A B, C D, being equal, the lines A B and CD must be parallel (Theo. XV. Bk. I.).

PROBLEM VIII. Two angles of a triangle being given, to find the third angle.

C

D

A

B

E

Draw the indefinite straight line ABE. At any point, B, make the angle ABC equal to one of the given angles (Prob. V.), and the angle CBD equal to the other given angle; then the angle DBE will be the third angle required. For these three angles are together equal to two right angles (Theo. I. Cor. 2, Bk. I.), as are also the three angles of every triangle (Theo. XIX. Bk. I.); and two of the angles at B having been made equal to two angles of the triangle, the remaining angle D B E must be equal to the third angle.

PROBLEM IX. Two sides of a triangle and the included angle being given, to construct the triangle.

C

Draw the straight line A B equal to one of the two given sides. At the point A make an angle, CA B, equal to the given angle (Prob. V.); and take AC equal to the other given side. Join B C, and the triangle A B C will be the one required (Theo. IV. Bk. I.). PROBLEM X. One side and two angles of a triangle

being given, to construct the triangle.

The two given angles will either be both adjacent to the given side, or one adjacent and the other opposite. In the latter case, find the third angle (Prob. VIII.); and

A

the two angles adjacent to the given side A4 will then be known.

с

B

B

In the former case, draw the straight line A B equal to the given side; at the point A, make an angle, BA C, equal to one of the adjacent angles, and at B an angle, A B C, equal to the other. Then the two sides A C, B C will meet, and form with AB the triangle required (Theo. V. Bk. I.).

PROBLEM XI. Two sides of a triangle and an angle opposite one of them being given, to construct the triangle.

Draw the indefinite straight line AB. At the point A make an angle BAC equal to the given angle, and make A C equal to that

C

side which is adjacent to the given A

angle. Then from C, as a center,

D

with a radius equal to the other side, describe an arc, which must either touch the line A B in D, or cut it in the points E and F, otherwise a triangle could not be formed.

When the arc touches A B, a straight line drawn from C to the point of contact, D, will be perpendicular to AB (Theo..