VIII. Cor. 2.) But by hypothesis AC-DF; and therefore AC AG (Ax. 1.) But the oblique line AC cannot be equal to AG, which lies nearer the perpendicular AB (Prop. XVII.); therefore, BC and EF cannot be unequal, and hence the angle A=C, and the angle BAC-F; and therefore, the triangles are equal.

PROPOSITION XX. THEOREM.

If two right lines which lie in the same plane are equidistant in two points, they are equidistant in all their points, and hence are parallel.

BG

Let AB and CD be two right lines ly- A E ing in the same plane, and let them be so related, that the lines EF and GH drawn from the points E and G in the line AB, perpendicular to CD, are equal each to each; and the lines AB and CD will be equidistant in all their parts.

CF

HD

For let the two lines AB and CD be brought into such position that the points E and G in the lihe AB may coincide. with the points F and H in the line CD, and (Prop. V.) the two lines will coincide in all their parts, and form but one line.

Now, as the line CD is a right line, and as all the points in the line AB coincide with corresponding points in the line CD when brought in the same position, if posited by the side of each other they will touch in all their parts, and the points E and G in the line AB will have the same relation to each of the other points in the line AB, as the points F and H in the line CD, to each of the other points in the line CD, and each point in the line AB will be similarly related to the corresponding points in the line CD, viz: they all touch each other. Now, if the line AB is moved, so that the points E and G may be equidistant from the points F and H, viz so that the line EF, joining the points E and F, may be equal. to the line GH joining the points G and H, the points E and G will still have the same relation to every other point in the line AB as before; and because the points E and G are similarly related to the points F and H, every other point in the AB is similarly related to its corresponding point in the line CD; that is, every point in the line AB is equidistant from the line CD, and hence the lines AB and CD are parallel.

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Scholium. It is evident that parallel lines would never meet each other, however far they may be produced; for ev

ery part by which the lines are produced, when added, becomes a part of the parallel lines, which, by definition, are equidistant in all their parts.

PROPOSITION XXI. THEOREM.

If two right lines are perpendicular to a third line, they are parallel to each other.

will they be parallel.

For, if it be possible, let the productions of the lines.

B

C

and

AC and BD, meet in some point O on either side of AB, if OCA and ODB are right lines and perpendicular to AB, there would be two perpendiculars, OA, OB, drawn from the same point on the same right line, which is impossible, (Prop. XVII. Cor. 1.)

PPOPOSITION XXII. THEOREM.

C

H

E

If a right line meet two other right lines, making the sum of the interior angles on the same side of the first mentioned line equal to two right angles, the two lines will be parallel. Let the line AB meet the two lines EC and BD, making the angles BAC, ABD, together equal to two right angles; then the lines EC and BD will Le parallel.

D

B

From G, the middle point of BA, draw the right line HGF, perpendicular to EC. It will also be perpendicular to BD. For, the sum BAC+ABD is equal to two right angles, by hypothesis; the sum BAC+ BAE is likewise equal to two right angles (Prop. III.); and taking away BAC from both, there will remain the angle ABD-BAE.

Again, the angles HGA, BGF, are equal (Prop. VI.); therefore, the triangles HGA and BGF, have each a side and two adjacent angles equal; therefore they are themselves equal, and the angle GHA is equal to the angle GFB: but GHA is a right angle by construction; therefore, GFB is a right angle; hence the two lines EC, BD, are perpendicular to the same straight line, and are therefore parallel (Prop. XXI.)

Scholium. When two parallel straight lines AB, CD, are met by a third line FE. the angles which are formed take. particular names.

Interior angles on the same side, are, those which lie within the parallels, and on the same side of the secant line: thus, OGB, BGE, are interior angles on the

F

same side; and so also are the angles OGA, GOC.

G

E

D

Alternate angles lie within the parallels, and on different sides of the secant line: AGO, DOG, are alternate angles; and so also are the angles COG, BGO.

Alternate exterior angles lie without the parallels, and on different sides of the secant line: EGB, COF, are alternate exterior angles; so also, are the angles AGE, FOD.

Opposite exterior and interior angles lie on the same side of the secant line, the one without and the other within the parallels, but not adjacent: thus, EGB, GOD, are opposite exterior and interior angles; and so also, are the angles AGE, GOC.

Cor. 1. If a straight line EF, meet two straight lines CD, AB, making the alternate angles AGO, GOD, equal to each other, the two lines will be parallel. For, to each add the angle OGB; we shall then have, AGO+OGE=GOD+ OGB; but AGO+0GB is equal to two right angles (Prop. III.); hence GOD+OGB is equal to two right angles: therefore, CD, AB, are parallel.

Cor. 2. If a straight line EF meet two straight lines CD, AB, making the exterior angle FGB equal to the interior and opposite angle GOD, the two lines will be parallel. For, to each add the angle OGB; we shall then have EGBOGB= GOD+OGB; but EGB+OGB is equal to two right angles; hence, GOD OGB is equal to two right angles; therefore, CD, AB, are parallel.

If a right line meet two parallel right lines, the sum of the interior angles on the same side will be equal to two right angles.

Let the parallels AB, CD, be met by the secant line FE: then will OGB+GOD, or OGA+GOC, be equal to two right angles.

For, if OGB+GOD be not equal to two right angles, let IGH be drawn, making the sum OGH+GOD equal to two right angles; then IH and CD will be parallel

(Prop. XXII.), and hence we shall have two lines GB, GH, drawn through the same point G and parallel to CD, which is impossible (Ax. 12.); hence, GB and GH should coincide, and OGB÷GOD is equal to two right angles. In the same manner it may be proved that OGA+COC is equal to two right angles.

Cor. 1. If OGB is a right angle, GOD will be a right angle also; therefore, every straight line perpendicular to one of two parallels, is perpendicular to the other.

Cor. 2. If a right line meet two parallel lines, the alternate angles will be equal.

Let AB, CD, (see diagram to the scholium of the preceding proposition), be the parallels, and FE the secant line. The sum OGB+GOD is equal to two right angles. But the sum OGB+OGA is also equal to two right angles (Prop. III.) Taking from each the angle OGB, and there remains OGA GOD. In the same manner we may prove that GOC =OGB.

Cor. 3. If a straight line meet two parallel lines, the opposite exterior and interior angles will be equal. For the sum OGB+GOD is equal to two right angles. But the sum OGB +DGB is also equal to two right angles. Taking from each the angle OGB, and there remains GOD=FGB. In the same manner we may prove that AGE=GOC.

Cor. 4. We see that of the eight angles formed by a line cutting two parallel lines obliquely, the four acute angles are equal to each other, and so also are the four obtuse angles.

PROPOSITION XXIV. THEOREM.

If two points in a right line be unequally distant from another right line, the former, by being produced on the side of the least distance, will continue to approach nearer and nearer to the latter, or its production, till at length it will meet it.

Let the two points F, G, in the CT tight line CD, be unequally distant from the right line AB, then the line CD, by being produced, shall ap- A

proach the line AB, or its prolongation, till it meets it. For the right lines AB and CD may be conceived to be generated by the motion of a point, and may be produced at pleasure by the continued motion of the point; and, (Def. S.). if a right line is traced by the motion of a point, that point moves always in the same direction. Now, if in the course of the production of two lines lying in the same plane, they are found to approach each other in any degree; or if the point G is found to be nearer to the line AB, than the point F; then, because the lines always preserve the same direction, the point D will be nearer the line AB than the point G, and so on for any production of the two lines; and hence, there must be a point more or less distant to which both of the lines tend, and in which they would meet if produced.

Cor. Hence, from the same point more than one parallel to a right line cannot be drawn.

PROPOSITION XXV. THEOREM.

Two right lines which are parallel to a third line are parallel to each other.

Let CD and AB be parallel to the third line EF; then are they parallel to each other.

C

Draw PQR perpendicular to EF, and cutting AB, CD. Since AB is parallel E to EF, PR will be perpendicular to AB (Prop. XXIII. Cor. 1.); and since CD is parallel to EF, PR will for a like reason be perpendicular to CD. Hence AB

and CD are perpendicular to the same straight line; hence they are parallel (Prop XXI.)

PROPOSITION XXVI. THEOREM.

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

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