H Let AB be the given straight line, C the os od 15. middle point, and ECF the perpendicular.qo od First, Since AC=CB, the two oblique lines up AD, DB, are equally distant from the perpendicular, and therefore equal (Prop. XV.). So, likewise, are the two oblique lines AE, EB, the A two AF, FB, and so on. Therefore every point in the perpendicular is equally distant from the extremities A and B. E oni Secondly, Let I be a point out of the perpendicular. If IA and IB be drawn, one of these lines will cut the perpendicular in D; from which, drawing DB, we shall have DB=DA. But the straight line IB is less than ID+DB, and ID+DB=ID+DA=IA; therefore, IB<IA; therefore, every point out of the perpendicular, is unequally distant from the extremities A and B. Cor. If a straight line have two points D and F, equally distant from the extremities A and B, it will be perpendicular to AB at the middle point C. PROPOSITION XVII. THEOREM. If two right angled triangles have the hypothenuse and a side of the one, equal to the hypothenuse and a side of the other, each to each, the remaining parts will also be equal, each to each, and the triangles themselves will be equal. In the two right angled triangles BAC, EDF, let the hypothenuse AC=DF, and the side BA=ED: then will the side BC=EF, the angle A=D, and the angle C-F. If the side BC is equal to EF, the like angles of the two triangles are equal (Prop. X.). Now, if it be possible, suppose these two sides to be unequal, and that BC is the greater. On BC take BG=EF, and draw AG. Then, in the two triangles BAG, DEF, the angles R and E are equal, being right angles, the side BA-ED by hypothesis, and the side BG=EF by construction: consequently, AG-DF (Prop. V. Cor.). 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. XV.); therefore, BC and EF cannot be unequal, and hence the angle A=D, and the angle C-F; and therefore, the triangles are equal (Prop. VI. Sch.). 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 C, on either side of AB, there would be two per B D 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. 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. From G, the middle point of BA, draw the straight line EGF, B perpendicular to EČ. 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. I.); and taking away BAC from both, there will remain the angle ABD=BAE. 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. 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 anglès OGA, GOC. D B 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+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. 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. right angles; then IH 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 is 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. Let AB, CD, 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. I.). Taking from each, the angle OGB, and there E D B 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 +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. 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 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. 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. 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 IH, CD, are not parallel, and will therefore meet if sufficiently produced. 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 parallel 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.). T Q D B PROPOSITION XXIII. THEOREM. Two parallels are every where equally distant. Two parallels AB, CD, being CH 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 G D. B perpendicular to CD (Prop. XX. Cor. 1.): and we are now to show that they will be equal to each other. 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. |