the angles B and E equal, because both are right angles; consequently, AG is equal to DF (P. V.) But, AC is equal to DF, by hypothesis: hence, AG and AC are equal, which is impossible (P. XV.). The hypothesis that BC and EF are unequal, is, therefore, absurd: hence, the triangles have all their sides equal, each to each, and are, consequently, equal in all of their parts; which was to be proved. PROPOSITION XVIII. THEOREM. If two straight lines are perpendicular to a third straight line, they will be parallel. Let the two lines AC, BD, be perpendicular to AB: then will they be parallel. For, if they could meet in a point 0, there would be two perpendiculars OA, OB, drawn from the same point to the same B D C straight line; which is impossible (P. XIV.): hence, the lines are parallel; which was to be proved. 1o. INTERIOR ANGLES ON THE SAME SIDE, are those that lie on the same side of the secant and within the other two lines. Thus, BGH and GHD are interior angles on the same side. 2o. EXTERIOR ANGLES ON THE SAME SIDE, are those that lie on the same side of the secant and without the other two lines. Thus, EGB and DHF are exterior angles on the same side. 3°. ALTERNATE ANGLES, are those that lie on opposite sides of the secant and within the other two lines, but not adjacent. Thus, AGH and GHD are alternate angles. E A -B C -D. H F 4°. ALTERNATE EXTERIOR ANGLES, are those that lie on opposite sides of the secant and without the other two lines. Thus, AGE and FHD are alternate exterior angles. 5°. OPPOSITE EXTERIOR AND INTERIOR ANGLES, are those that lie on the same side of the secant, the one within and the other without the other two lines, but not adjacent. Thus, EGB and GHD are opposite exterior and interior angles. If two straight lines meet a third straight line, making the sum of the interior angles on the same side equal to two right angles, the two lines will be parallel. Let the lines KC and IID meet the line BA, making the sum of the angles BAC and ABD equal to two right angles then will KC and HD be parallel. angles (P. I.); the sum of the angles FAG and GBD is equal to two right angles, by hypothesis: hence (A. 1), GBE+ GBD = FAG + GBD. Taking from both the common part GBD, we have the angle GBE equal to the angle FAG. Again, the angles BGE and AGF are equal, because they are vertical angies (P. II.): hence, the triangles GEB and GFA have two of their angles and the included side equal, each to each; they are, therefore, equal in all their parts (P. VI.): hence, the angle GEB is equal to the angle GFA. But, GFA is a right angle, by construction; GEB must, therefore, be a right angle hence, the lines KC and HD are both perpendicular to EF, and are, therefore, parallel (P. XVIII.); which was to be proved. Cor. 1. If two straight lines are cut by a third straight line, making the alternate angles equal to each other, the two straight lines will be parallel. the second sum is also equal to two right angles; therefore, from what has just been shown, AB and CD are parallel. Cor. 2. If two straight lines are cut by a third, making the opposite exterior and interior angles equal, the two straight lines will be parallel. Let the angles EGB and GHD be equal: Now EGB and AGH are equal, because they are vertical angles (P. II.); and consequently, AGH and GHD are equal: hence, from Cor. 1, AB and CD are parallel. PROPOSITION XX. THEOREM. If a straight line intersect two parallel straight lines, the sum of the interior angles on the same side will be equal to two right angles. Let the parallels AB, CD, be cut by the secant line FE: then will the sum of HGB and GHD be equal to two right angles. For, if the sum of HGB and GHD is not equal to two right angles, let IGL be drawn, making the sum of HGL and GHD equal to two right angles; then IL and CD will A C E -B L -D Ή F be parallel (P. XIX.); and consequently, we shall have two lines GB, GL, drawn through the same point G and parallel to CD, which is impossible (A. 13): hence, the sum of HGB and GHD, is equal to two right angles; which was to be proved. In like manner, it may be proved that the sum of HGA and GHC, is equal to two right angles. Cor. 1. If HGB is a right angle, GHD will be a right angle also hence, if a line is perpendicular to one of two parallels, it is perpendicular to the other also. Cor. 2. If a straight line meet twc parallels, the alternate are equal. Taking away the common part BGH, there re- mains the angle GHD equal to HGA. it may be shown that BGH and GHC are equal. Cor. 3. If a straight line meet two parallels, the opposite exterior and interior angles will be equal. The angles DHG and HGA are equal, from what has just been shown. The angles HGA and BGE are equal, because they are vertical hence, DHG and BGE are equal. In like manner, it may be shown that CHG and AGE are equal. Scholium. Of the eight angles formed by a line cutting two parallel lines obliquely, the four acute angles are equal, and so, also, are the four obtuse angles. PROPOSITION XXI. THEOREM. If two straight lines intersect a third straight line, 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 two lines CD, IL, meet the line EF, making the sum of the interior angles HGL, GHD, less than two right angles then will IL and CD meet if sufficiently produced. For, if they do not meet, they must be parallel (D. 16). But, if they were parallel, the sum of the interior angles HGL, GHD, would be equal to two right angles (P. XX.), which is contrary to the hypothesis: hence, E G C -D H F IL, CD, will meet if sufficiently produced; which was to be proved. |