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drawn through the point G, except the line CD, can be parallel to AB.

Cor. Those lines which are perpendicular to the same line, are parallel to each other.

(40.) The principle contained in the Corollary, that lines which are perpendicular to the same line are parallel to each other, has a practical application in the use of the instrument called the T-square. It consists of two straight rulers fixed at right-angles to each other, forming a sort of double square. A straight line being drawn in a direction perpendicular to that in which the parallels are required to be drawn, the

cross-piece of the T-square is laid upon this line, and the piece at right-angles to it gives the direction of the parallels. The ruler being moved along the paper, keeping the cross-piece coincident with the line first drawn, any number of parallel lines may thus be formed.

By means of a movable joint, the rulers which compose this kind of square can be made to form any angle whatever with each other, which, in practice, is found to be quite convenient. To suit this more general case, the preceding corollary might read "Those lines which make the same angle with the same line, are parallel."

PROPOSITION XIX.

PROBLEM. Through a given point D, to draw a line parallel to a given line AB.

Draw any line DC, meeting AB at C; and with C and D as centres, and CD as a radius, describe the arcs (Post. III,) DB, CH, the former meeting AB at B

F

A

C

D

B

Then with C as a centre, and with a radius

equal to the distance from B to D, describe an arc to meet the arc CH at the point F. The line DF, being drawn, will be parallel to AB.

For (Prop. vIII,) the alternate angles FDC, DCB are equal; and therefore (Prop. XVIII,) the line FD is parallel to AB.

PROPOSITION XX.

THEOREM. When a straight line cuts two parallel lines, the exterior angle is equal to the interior and opposite one, on the same side; and the two interior angles, on the same side, are together equal to two right-angles.

Let the line FG cut the two parallel lines AB, CD; then will the exterior angle FHB be equal to the interior and opposite angle HKD, on the same side of the line FG; and the two interior angles BHK, DKH,

F

H

A

B

C

K

D

G

taken together, will be equal to two right angles. For, since the two lines AB, CD are parallel, the angle AHK is equal to the alternate angle DKH (Prop. XVII;) but the angle AHK is equal to the opposite angle FHB, (Prop. II ;) therefore the angle FHB is equal to the angle HKD, (Ax. I.)

Again, because the two adjacent angles BHF, BHK are together equal to two right-angles, (Prop. 1,) of which the angle FHB has been shown to equal the angle HKD; therefore the two angles BHK, DKH, taken together, are also equal to two right-angles.

Cor. 1. And, conversely, if one line meet two other

lines, making the exterior angle equal to the interior and opposite one, those two lines will be parallels.

Cor. 2. If a line, cutting two other lines, make the sum of the two interior angles, on the same side, less than two right-angles, those two lines will not be parallel, and consequently will meet each other when produced.

PROPOSITION XXI.

THEOREM. Those lines which are parallel to the same line, are parallel to each other.

Let the lines AB, CD be each parallel to FG; then will the lines AB, CD be parallel to each other.

For, let the line HL be per

A

с

H

K

B

D

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pendicular to FG; then will this line be also perpendicular to both the lines AB, CD, (Prop. xvII, Cor.) and consequently the two lines AB, CD are parallels, (Prop. XVIII, Cor.)

PROPOSITION XXII.

THECREM. If two angles have their sides parallel, and lying in the same direction, the two angles will be equal.

Let BAC and FDG be the two angles, having AB parallel to DF, and AC parallel to DG; then will the angles be equal.

For, produce CA, if necessary, till it meets DF in H

D

A

C

Then, since AB is parallel to DF, the angle BAC is

equal to FHA, (Prop. xx ;) and since HC is parallel to DG, the angle FHA is equal to FDG, (Prop. xx ;) hence the angle BAC is equal to the angle FDG, (Ax. I.)

THEOREM.

PROPOSITION XXIII.

When each side of a polygonal figure is produced, in the same direction, the sum of all the exterior angles will be equal to four right-angles.

For, if from any point in the same plane, straight lines be drawn parallel to the sides of the figure, the angles contained by the straight lines about that point will be equal to the exterior angles of the figure (Prop. XXII,) each to each, because

B

A

their sides are parallel to the sides of the figure. Thus the angles a, b, c, etc., are respectively equal to the exterior angles A, B, C, etc.; but the former angles are together equal to four right-angles, (Prop. 1, Cor. 3;) therefore all the exterior angles of the figure are together equal to four right-angles.

A convex

Scholium. This proposition must be restricted to the case in which the polygon is strictly convex. polygon may be defined to be one such that no side, by being produced in either direction, can divide the polygon. The polygon ABCDFG is not convex, since it may be divided by producing either of the sides CD or FD.

A

B

This polygon is said to have a re-entering angle at D.

THEOREM.

PROPOSITION XXIV.

In any convex polygon, the sum of all the interior angles, taken together, is equal to twice as many right-angles as the polygon has sides, wanting four right-angles.

F/f

a

B

C

Let ABCDFG be a convex polygon. Conceive the sides to be produced all in the same direction, forming exterior angles, which we will denote by the capital letters A, B, C, etc., while their corresponding interior angles are denoted by the small letters a, b, c, etc. Now any exterior angle, together with its adjacent interior angle, as A+a, is equal to two right-angles, (Prop. 1;) therefore the sum of all the interior angles, together with all the exterior angles, is equal to twice as many right-angles as the polygon has sides; but the sum of all the exterior angles is equal to four right-angles, (Prop. XXIII;) therefore the sum of all the interior angles is equal to twice as many right-angles as the polygon has sides, wanting four rightangles.

Cor. 1. In any triangle, the sum of all the three angles is equal to two right-angles.

Cor. 2. In any quadrilateral, the sum of all the four interior angles is equal to four right-angles.

(41.) In order that this proposition may hold good in polygons having re-entering angles, as in the case of the polygon ABCDFG (next page,) which has a re-entering angle at D, we must take, for the interior angle at D, the angle which remains after subtracting

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