III.); hence the angles CA B, E FG, measured by these arcs, are also equal (Prop. V. Bk. III.).

PROBLEM VI.

300. 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 cen

tre C draw the line CD perpendicular

to AB (Prob. III.); it will bisect the A arc A D B in the point D.

For CD being a radius perpendicu

C

B

D

lar to a chord AB, must bisect the arc ADB which is subtended by that chord (Prop. VI. Bk. III.).

Secondly. Let ACB be the angle which it is required to bisect. From C as a centre, 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 ACB.

For the angles ACD, BCD are equal, being measured by the equal arcs AD, D B (Prop. V. Bk. III.).

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

PROBLEM VII.

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

Let A be the given point, and

CD the given straight line.

From A draw a straight line, AE, to any point, E, in CD.

A

C

Then draw A B, making the angle EAB equal

E

B

angle AEC (Prob. V.); and AB is parallel to CD.

For the alternate angles EA B, A E C, made by the straight line

AE meeting the two straight lines A B, CD, being equal, the lines AB and CD must be parallel (Prop. XX. Bk. I.).

PROBLEM VIII.

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

C

D

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 C B D equal to the other given angle; then the angle D BE will be the third angle required.

A

B

E

For these three angles are together equal to two right angles (Prop. I. Cor. 2, Bk. I.), as are also the three angles of every triangle (Prop. XXVIII. Bk. I.); and two of the angles at B having been made equal to two angles of the triangle, the remaining angle DBE must be equal to the third angle.

PROBLEM IX.

304. 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 A C equal to the other given side. Join BC; and the triangle ABC will be the one required. (Prop. V. Bk. I.).

Α'

B

PROBLEM X.

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

A4

C

B

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

PROBLEM XI.

306. 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 side which is adjacent to the given angle.

A

E

C

B

Then from C, as a centre, 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 A B (Prop. XI. Bk. III.), and the right-angled triangle CAD will be the triangle required.

When the arc cuts A B in two points, E and F, lyi

on the same side of the point A, draw the straight lines CE, CF; and each of the two triangles CA E, CAF will satisfy the conditions of the problem. If, however, the two points E and F should lie on different sides of the point A, only one of the triangles, as CAF, will satisfy all the conditions; hence that will be the triangle required.

307. Scholium. The problem would be impossible, if the side opposite the given angle were less than the perpendicular let fall from the point C on the straight line AB.

PROBLEM XII.

308. The three sides of a triangle being given, to construct the triangle.

C

Draw the straight line A B equal to one of the given sides; from the point A as a centre, with a radius equal to either of the other two sides, describe an arc; from the point B, with a radius equal to the third side, describe another arc cutting the former in the point C; draw the straight lines AC, BC; and the triangle ABC will be the one required (Prop. XVIII. Bk. I.).

A

B

309. Scholium. The problem would be impossible, if one of the given sides were greater than the sum of the other two.

PROBLEM XIII.

D

310. Two adjacent sides of a parallelogram and the included angle being given, to construct the parallelogram. Draw the straight line A B equal to one of the given sides. At the point A make an angle, BAD, equal to the given angle, and take AD equal to the other given side. From A

B

the point D, with a radius equal to A B, describe an arc; and from the point B as a centre, with a radius equal to A D, describe another arc cutting the former in the point C. Draw the straight lines CD, CB; and the parallelogram ABCD will be the one required.

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

311. Cor. If the given angle is a right angle, the figure will be a rectangle; and if the adjacent sides are also equal, the figure will be a square.

PROBLEM XIV.

312. A circumference, or an arc, being given, to find the centre of the circle.

Take any three points, A, B, C, on the given circumference, or arc. Draw the chords AB, BC, and bisect them by the perpendiculars DE and FE (Prob. I.); the point E, in which these perpendiculars meet, is the centre required.

E

D

B

C

For the perpendiculars DE, FE must both pass through the centre (Prop. VI. Cor. 2, Bk. III.), and E being the only point through which they E must be the centre.

both pass,

313. Scholium. By the same construction, a circumference may be made to pass through three given points, A, B, C, not in the same straight line; and also a circumference described in which a given triangle, ABC, shall be inscribed.

PROBLEM XV.

314. Through a given point to draw a tangent to a given circle.