Secondly. Let us suppose BC=BE; then the triangle CAB will be equal to the triangle BAE; for BC=BE, the side AB is common, and the angle CBA=ABE; hence, the sides AC and AE are equal (P. 5, c): therefore, two oblique lines, which meet the given line at equal distances from the perpendicular, are equal. Thirdly. Since the point C is within the triangle FDA, the sum of the sides FD, DA, is greater than the sum of the lines FC, CA (P. 8): therefore AD, the half of the broken line FDA, is greater than AC, the half of FCA: consequently, the oblique line which cuts off the greater distance, is the longer. Cor. 1. The perpendicular measures the shortest distance of a point from a line. Cor. 2. From the same point to the same straight line, only two equal straight lines can be drawn; for, if there could be more, we should have at least two equal oblique lines on the same side of the perpendicular, which is impossible. PROPOSITION XVI. THEOREM. If at the middle point of a straight line, a perpendicular to this line be drawn: 1st. Any point of the perpendicular will be equally distant from the extremities of the line: 2d. Any point, without the perpendicular, will be unequally distant from the extremities. Let AB be the given straight line, C its middle point, and ECF the perpendicular. First. Let D be any point of the perpendicular, and draw DA and DB. Then, since AC CB, the two oblique lines AD, DB, are equal (P. 15). the two oblique lines, So, likewise, are AF, FB, and so on. Therefore, any F D B E Secondly. Let I be any point out of the perpendicular. If IA and IB be drawn, one of these lines will cut the perpendicular in some point as D; from this point, drawing DB, we shall have DB=DA. For, the straight line IB is less than ID+DB, and ID+DB=ID+DA=IA; therefore, IBIA; consequently, any point out of the perpendicular, is unequally distant from the extremities A and B. Cor. Conversely: if a straight line have two points E and F, each of which is 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 triangles are equal. Let BAC and EDF be two right-angled triangles, having the hypothenuse AC-DF, and the side BA=ED: then will the triangle BAC be equal to the triangle EDF. On BC take BG=EF, and draw AG. Then, in the two triangles BAG, EDF, the angles B and E are equal, being right angles, the side BA=ED by hypothesis, and the side, BG=EF by construction; consequently, AG=DF (P. 5, c). But by hypothesis AC=DF; and therefore, AC=AG (A. 1). But the oblique line AC cannot be equal to AG, since BC is greater than BG (P. 15); consequently, BC and EF cannot be unequal, and hence, the triangles are equal (P. 10). PROPOSITION XVIII. THEOREM. If two straight lines are perpendicular to a third line, they are parallel to each other. Let the two lines AC, BD, be perpendicular to AB; then will they be parallel. If two straight lines meet a third line, making the sum of the interior angles on the same side equal to two right angles, the two lines are parallel. Let the two lines KC, HD, meet the line BA, making the angles BAC, ABD, together equal to two right angles: then the lines KC, HD, will be parallel. From G, the middle point of BA, draw the straight line EGF, perpendicular to KC: then, it will also be perpendicular to HD. For, the sum BAC+ABD is E B H -D G K C AF equal to two right angles, by hypothesis; the sum ABD+ABE is likewise equal to two right angles (P. 1): taking away ABD from both, there will remain the angle BAC-ABE. Again, the angles EGB, AGF, are equal (P. 4); there fore, the triangles EGB and AGF, have each a side and two adjacent angles equal; therefore, they are themselves equal, and the angle GEB is equal to the angle GFA (P. 6, c). But GEB is a right angle by construction; therefore, GFA is a right angle; hence, the two lines KC, HD, are perpendicular to the same straight line, and are therefore parallel (P. 18). 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 E A G -B -D H F side of the secant line; thus, HGB, GHD, are interior angles on the same side; and so also are the angles HGA, GHO Alternate angles lie within the parallels, and on different sides of the secant line, but not adjacent; AGH, GHD, are alternate angles; and so also are the angles GHC, BGH. Alternate exterior angles lie without the parallels, and on different sides of the secant line, but not adjacent: EGB, CHF, are alternate exterior angles; so also are the angles AGE, FHD. 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, GHD, are opposite exterior and interior angles; and so also, are the angles AGE, GHC. Cor. 1. If two straight lines meet a third line, making the alternate angles equal, the straight lines are parallel. Let the straight line EF meet the two straight lines CD, AB, making the alternate angles AGH, GHD, equal to each other: then will AB and CD be parallel, For, to each of the equal angles, add the angle HGB; we shall then have AGH+HGB=GHD+HGB. But AGH+HGB is equal to two right angles (P. 1): hence, GHD+HGB is also equal to two right angles (A. 1): then CD and AB are parallel (P. 19.) E A C -D H -B 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 GHD, the two lines will be parallel. For, to each add the angle HGB: we shall then have, EGB+HGB=GHD+HGB: but EGB+HGB is equal to two right angles; hence, GHD+HGB is equal to two right angles; therefore, CD, AB, are parallel (P. 19). 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. Let the parallels AB, CD, be met by the secant line FE: then will HGB+GHD, or HGA+GHC, be equal to two right angles. For, if HGB+GHD_be not equal to two right angles, let IGL be drawn, making the sum HGL+ A GHD equal to two right an E I -B. L gles; then IL and CD will be parallel (P. 19); and hence, H F we shall have two lines GB, GL, drawn through the same point G and parallel to CD, which is impossible (A. 13): hence, HGB+GHD is equal to two right angles. In the same manner it may be proved that HGA+GHC is equal to two right angles. Cor. 1. If HGB is a right angle, GHD 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 straight lines, the alternate angles will be equal. Let AB, CD, be two parallels, and FE the secant line. |