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PROPOSITION XVIII. THEOREM.

If two straight lines are perpendicular to a third line, they will be parallel to each other: in other words, they will never meet, how far soever either way, both of them be produced.

Let the two lines AC, BD, A be perpendicular to AB; then will they be parallel.

For, if they could meet in a point O, on either side of AB, there would be two per

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pendiculars OA, OB, let fall from the same point on the same straight line; which is impossible (Prop. XIV.).

PROPOSITION XIX. THEOREM.

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

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Let the two lines EC, BD, meet the third line BA, making the angles BAC, ABD, together equal to two right angles: then the lines EC, BD, will be parallel.

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From G, the middle point of BA, draw the straight line EGF, 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 kewise equal to two right angles (Prop. I.); and taking away BAC from both, there will remain the angle ABD=BAE.

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Again, the angles EGA, BGF, are equal (Prop. IV.); therefore, the triangles EGA and BGF, have each a side and two adjacent angles equal; therefore, they are themselves equal, and the angle GEA is equal to the angle GFB (Prop. VI. Cor.): but GEA 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. XVIII.).

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

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Interior angles on the same side, are those which lie within the parallels, and on the same side of the secant line: thus, OGB, GOD, are interior angles on the same side; and so also are the the angles OGA, GOC.

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.

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

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+OGB=GOD+OGB; but AGO+OGB is equal to two right angles (Prop. I.); 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 EGB 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 EGB+OGB=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.

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PROPOSITION XX. THEOREM.

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

For, if OGB+GOD be not equal to two right angles, let IGH be drawn, making the sum C OGH+GOD equal to two

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right angles; then III and CD will be parallel (Prop. XIX.), 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 18 equal to two right angles. In the same manner it may be proved that OGA+GOC 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 straight line meet two parallel lines, the alternate angles will be equal.

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Let AB, CD, be the parallels, and A 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. I.). Taking from each, the angle OGB, and there remains GA=GOD. In the same manner we may prove that GOC=OGB.

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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 +EGB is also equal to two right angles. Taking from each the angle OGB, and there remains GOD=EGB. In the same manner we may prove that AGE=GOC.

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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.

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PROPOSITION XXI. THEOREM.

If a straight line meet two other straight lines, making the sum of the interior angles on the same side less than two right angles, the two lines will meet if sufficiently produced.

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Let the line EF meet the two lines CD, IH, making the sum of the interior angles OGH, GOD, less than two right angles: then will IH and CD meet if sufficiently produced.

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For, if they do not meet they are parallel (Def.12.). But they are not parallel, for if they were, the sum of the interior angles OGH, GOD, would be equal to two right angles (Prop. XX.), whereas it is less by hypothesis: hence, the lines ÌH, CD, are not parallel, and will therefore meet if sufficiently produced.

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Cor. It is evident that the two lines IH, CD, will meet on that side of EF on which the sum of the two angles OGH, GOD, is less than two right angles

PROPOSITION XXII. THEOREM.

Two straight lines which are parallel to a third line, are paralle to each other.

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

Draw PQR perpendicular to EF, and cutting AB, CD. Since AB is parallel to EF, PR will be perpendicular to AB (Prop.E XX. Cor. 1.); and since CD is parallel to EF, PR will for a like reason be perpen-C dicular to CD. Hence AB and CD are. perpendicular to the same straight line; hence they are parallel (Prop. XVIII.).

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PROPOSITION XXIII. THEOREM.

Two parallels are every where equally distant.

Two parallels AB, CD, being C H given, if through two points E and F, assumed at pleasure, the straight lines EG, FH, be drawn perpendicular to AB,these straight A lines will at the same time be perpendicular to CD (Prop. XX. Cor. 1.): and we are now to show that they will be equal to each other.

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If GF be drawn, the angles GFE, FGH, considered in reference to the parallels AB, CD, will be alternate angles, and therefore equal to each other (Prop. XX. Cor. 2.). Also, the straight lines EG, FH, being perpendicular to the same straight line AB, are parallel (Prop. XVIII.); and the angles EGF, GFH, considered in reference to the parallels EG, FH, will be alternate angles, and therefore equal. Hence the two triangles EFG, FGH, have a common side, and two adjacent angles in each equal; hence these triangles are equal (Prop. VI.) ; therefore, the side EG, which measures the distance of the parallels AB and CD at the point E, is equal to the side FH, which measures the distance of the same parallels at the point F.

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PROPOSITION XXIV. THEOREM.

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

Let BAC and DEF be the two angles, having AB parallel to ED, and AC to EF; then will the angles be equal.

PROPOSITION XXV. THEOREM.

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For, produce DE, if necessary, till it meets AC in G. Then, since EF is parallel to GC, the angle DEF is equal to DGC (Prop. XX. Cor. 3.); and since DG is parallel to AB, the angle DGC is equal to BAC; hence the angle DEF is equal to BAC (Ax. 1.).

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Scholium. The restriction of this proposition to the case where the side EF lies in the same direction with AC, and ED in the same direction with AB, is necessary, because if FE were produced towards H, the angle DEH would have its sides parallel to those of the angle BAC, but would not be equal to it. In that case, DEH and BAC would be together equal to two right angles. For, DEH+DEF is equal to two right angles (Prop. I.); but DEF is equal to BAC: hence, DEH + BAC is equal to two right angles.

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In every triangle the sum of the three angles is equal to two right angles.

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Let ABC be any triangle: then will the angle C+A+B be equal to two right angles. For, produce the side CA towards D, and at the point A, draw AE parallel to BC. Then, since AE, CB, are parallel, and CAD cuts them, the exterior angle DAE will be equal to its inte-C rior opposite one ACB (Prop. XX. Cor. 3.); in like manner, since AE, CB, are parallel, and AB cuts them, the alternate angles ABC, BAE, will be equal: hence the three angles of the triangle ABC make up the same sum as the three angles CAB, BAE, EAD; hence, the sum of the three angles is equal to two right angles (Prop. I.).

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Cor. 1. Two angles of a triangle being given, or merely their sum, the third will be found by subtracting that sum from two right angles.

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