BOOK V.

PROBLEMS.

Postulates.

1. A straight line may be drawn from any one point to any other point.

2. A terminated straight line may be produced to any length in a straight line.

3. From the greater of two straight lines, a part may be cut off equal to the less.

4. A circumference may be described from any center; and with any radius.

PROBLEM I.

To bisect a given straight line.

Let AB be the given straight line which it is required to bisect.

From the center A, with a radius great

er than the half of AB, describe an arc of A-
a circle (Postulate 4); and from the cen-
ter B, with the same radius, describe an-
other arc intersecting the former in D and

B

E. Through the points of intersection, draw the straight line DE (Post. 1); it will bisect AB in C.

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 point of AB (Prop. XVIII. Cor., B. I.). Therefore the line DE divides the line AB into two equal parts at the point C.

PROBLEM II.

To draw a perpendicular to a straight line, from a given point in that line.

Let BC be the given straight line, and A the point given in it; it is required to draw a straight line perpendicular to BC through the given point A.

B

A

In the straight line BC take any point B, and make AC equal to AB (Post. 3). From B as a center, with a radius greater than BA, describe an arc of a circle (Post. 4); and from C as a center, with the same radius, describe another arc intersecting the former in D. Draw AD (Post. 1), and it will be the perpendicular required.

For, the points A and D, being equally distant from B and C, must be in a line perpendicular to the middle of BC (Prop. XVIII., Cor., B. I.). Therefore AD has been drawn perpendicular to BC from the point A.

Scholium. The same construction serves to make a right angle BAD at a given point A, on a given line BC.

PROBLEM III.

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

Let BD be a straight line of unlimited length, and let A be a given point without it. It is required to draw a perpendicular to BD from the point A.

Take any point E upon the other side of BD; and from the center A, with the B radius AE, describe the arc BD cutting the line BCD in the two points B and D. From the points B and D as centers, de

scribe two arcs, as in Prob. II., cutting each other in F. Join AF, and it will be the perpendicular required.

For the two points A and F are each equally distant from the points B and D; therefore the line AF has been drawn perpendicular to BD (Prop. XVIII., Cor., B. I.), from the given point A.

PROBLEM IV.

At a given point in a straight line, to make an angle cqua ts a given angle.

Let AB be the given straight line, A the given point in it, and C the given angle; it is required to make an angle at the point A in the straight line AB, that shal! A be equal to the given angle C.

BC

D

With C as a center, and any radius, describe an arc DE terminating in the sides of the angle; and from the point A as a center, with the same radius, describe the indefinite arc BF. Draw the chord DE; and from B as a center, with a radius equal to DE, describe an arc cutting the arc BF in G. Draw AG, and the angle BAG will be equal to the given angle C.

For the two arcs BG, DE are described with equal radii, and they have equal chords; they are, therefore, equal (Prop. III., B. III.). But equal arcs subtend equal angles (Prop IV., B. III.); and hence the angle A has been made equal to the given angle C.

PROBLEM V.

To bisect a given arc or angle.

First. Let ADB be the given arc which it is required to bisect.

Draw the chord AB, and from the center C draw CD perpendicular to AB (Prob. III.); it will bisect the arc ADB (Prop. VI., B. III.), because CD is a radius perpendicular to a chord.

D

B

Secondly. Let ACB be an angle which it is required to bisect. From C as a center, with any radius, describe an arc AB; and, by the first case, draw the line CD bisecting the arc ADB. The line CD will also bisect the angle ACB. For the angles ACD, BCD are equal, being subtended by the equal arcs AD, DB (Prop. IV., B. III.).

Scholium. By the same construction, each of the halves AD, DB may be bisected; and thus by successive bisections an arc or angle may be divide into four equal parts, intc eight, sixteen, &c.

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

Let A be the given point, and BC the given straight line; it is required to draw through the point A, a straight line parallel to BC.

B

A

E

In BC take any point D, and join AD. Then at the point A, in the straight line AD, make the angle DAE equal to the angle ADB (Prob. IV.).

Now, because the straight line AD, which meets the two straight lines BC, AE, makes the alternate angles ADB, DAE equal to each other, AE is parallel to BC (Prop. XXII., B. I.). Therefore the straight line AE has been drawn through the point A, parallel to the given line BC.

PROBLEM VII.

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

D

B

E

C

The three angles of every triangle are together equal to two right angles (Prop. XXVII., B. I.). Therefore, draw the indefinite line ABC. At the point B make the angle ABD equal to one of the given angles (Prob. IV.), and the angle DBE equal to the other given angle; then will the angle EBC be equal to the third angle of the triangle. For the three angles ABD, DBE, EBC are together equal to two right angles (Prop. II., R I.), which is the sum of all the angles of the triangle.

PROBLEM VIII.

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

Draw the straight line BC equal to one of the given sides. At the point B make the angle ABC equal to the given angle (Prob. IV.); and take AB equal to the other given side.

Join AC. and ABC will be the

A

C

triangle required. For its sides AB, BC are made equal to the given sides, and the included angle B is made equal to the given angle.

PROBLEM IX.

Given one side and two angles of a triangle, 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. VII.); and then the two adjacent angles will be known.

D

E

Draw the straight line AB equal to the given side; at the point A make the angle BAC equal to one of the adjacent angles; and at the point B make the angle ABD equal to the other adjacent angle. The two lines AC, BD will cut each other in E, and ABE will be the triangle required; for its side AB is equal to the given side, and two of its angles are equal to the given angles.

A

B

PROBLEM X.

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

Draw the straight line BC equal to one of the given sides. From the point B as a center, with a radius equal to one of the other sides, describe an arc of circle; and from the point C as a center, with a radius equal to the third side, describe another arc cutting the former in A. Draw AB, AC; then will B

ABC be the triangle required, because its three sides are equal to the three given straight lines.

Scholium. If one of the given lines was greater than the sum of the other two, the arcs would not intersect each other, and the problem would be impossible; but the solution will always be possible when the sum of any two sides is greater than the third.