PROBLEM IV.

At a given point in a straight line, to make an angle equa 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 shall A be equal to the given angle C.

F

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.

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 divided into four equal parts, intc eight, sixteen, &c.

PROBLEM VI.

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.

A

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 ADB, 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 B

A

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

B

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.

PROBLEM XI.

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

E

A

Draw an indefinite straight line BC. At the point B make the angle ABC equal to the given angle, and make BA equal to that side which is adjacent to the given angle. Then from A as a center, with a radius B equal to the other side, describe an arc cutting BC in the points E and F. Join AE, AF. If the points E and F both fall on the same side of the angle B, each of the triangles ABE, ABF will satisfy the given conditions; but if they fall upon different sides of B, only one of them, as ABF, will satisfy the conditions, and therefore this will be the triangle required.

If the points E and F coincide with one another, which will happen when AEB is a right angle, there will be only one triangle ABD, which is the triangle required.

Scholium. If the side opposite the given angle were less than the perpendicular let fall from A upon BC, the problem would be impossible.

PROBLEM XII.

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

C

B

Draw the straight line AB equal to one of the given sides. At the point A make the angle BAC equal to the given angle; and take AC equal to the other given side. From the point C as a center, with a radius equal to AB, describe an arc; and from the point B as a center, with a radius equal to AC, describe another arc intersecting the former in D. Draw BD, CD; then will ABDC be the parallelogram required.

A

For, by construction, the opposite sides are equal; therefore the figure is a parallelogram (Prop. XXX., B. I.), and it is formed with the given sides and the given angle

Cor. If the given angle is a rigat angle, the figure will be a rectangle; and if, at the same time, the sides are equal, it will be a square.

PROBLEM XIII.

To find the center of a given circle or arc.

Let ABC be the given circle or arc; it is required to find its center.

D

Take any three points in the aïe, as A B, C, and join AB, BC. Bisect AB in D (Prob. I.), and through D draw DF A perpendicular to AB (Prob. II.). In the same manner, draw EF perpendicular to BC at its middle point. The perpendiculars DF, EF will meet in a point F equally distant from the points A, B, and C (Prop. VII., B. III.); and therefore Fis the center of the circle.

B

Scholium. By the same construction, a circumference may be made to pass through three given points A, B, C; and also, a circle may be described about a triangle.

PROBLEM XIV.

Through a given point, to draw a tangent to a given circle

First. Let the given point A be without the circle BDE; it is required to draw a tangent to the circle through the point A.

Find the center of the circle C, and join AC. Bisect AC in D; and with D as a center, and a radius equal to

E

B

D

AD, describe a circumference intersecting the given circumference in B. Draw AB, and it will be the tangent required.

Draw the radius CB. The angle ABC, being inscribed in a semicircle, is a right angle (Prop. XV., Cor. 2, B. III.). Hence the line AB is a perpendicular at the extremity of the radius CB; it is, therefore, a tangent to the circumference (Prop IX., B. III.).

Secondly. If the given point is in the circumference of the circle, as the point B, draw the radius BC, and make BA perpendicular to BC. BA will be the tangent required (Prop. IX., B. III.).