PROPOSITION XV. THEOREM. If from a point without a straight line a perpendicular be let fall on the line, and oblique lines be drawn to different points of it: 1o. The perpendicular will be shorter than any oblique line: 2o. Any two oblique lines that meet the given line at points equally distant from the foot of the perpendicular, will be equal: Of two oblique lines that meet the given line at points unequally distant from the foot of the perpendicular, the one which meets it at the greater distance will be the longer. Let A be be a given point, DE a given straight line, AB a perpendicular to DE, and AD, AC, AE oblique lines, BC being equal to BE, and BD greater than BC. Then will AB be less than any of the oblique lines, AC will be equal to AE, and AD greater than AC. Prolong AB until BF is equal to AB, and draw FC, FD. 1o. In the triangles ABC, FBC, we have the side AB equal to BF, by construction, the side BC common, and the included angles ABC and FBC equal, because both are right angles: hence, FC is equal to AC (P. V.). But, AF is shorter than ACF (A. 12): hence, AB, the half of AF, is shorter than AC, the half of ACF; which was to be proved. 2o. In the triangles ABC and ABE, we have the side BC equal to BE, by hypothesis, the side AB common, and the included angles ABC and ABE equal, because both are right angles: hence, AC is equal to AE; which was to be proved. 3o. It may be shown, as in the first case, that AD is equal to DF. Then, because the point C lies within the triangle ADF, the sum of the lines AD and DF will be greater than the sum of the lines AC and CF (P. VIII.): hence, AD, the half of ADF, is greater than AC, the half of ACF; which was to be proved. Cor. 1. The perpendicular is the shortest distance from a point to a line. Cor. 2. From a given point to a given straight line, only two equal straight lines can be drawn; for, if there could be more, there would be at least two equal oblique lines on the same side of the perpendicular; which is impossible. If a perpendicular be drawn to a given straight line at its middle point: 1o. Any point of the perpendicular will be equally distant from the extremities of the line: 2o. Any point, without the perpendicular, will be unequally distant from the extremities. Let AB be a given straight line, C its middle point, and EF the perpendicular. Then will any point of EF be equally distant from A and B; and any point without EF, will be unequally distant from A and B. 1°. From any point of EF, as D, draw the lines DA and DB. Then will DA and DB be equal (P. XV.): hence, D is A D F E # B equally distant from A and B ; which was to be proved. D F 2o. From any point without EF, as I, draw IA and IB. One of these lines, as IA, will cut EF in some point D; draw DB. Then, from what has just been shown, DA and DB will be equal; but IB is less than the sum of ID and DB (P. VII.); and because the sum of ID and DB is equal to the sum of ID and DA, or IA, we have IB less than IA: hence, I is unequally distant from A and B; which was to be proved. A E B Cor. If a straight line EF have two of its points E and F equally distant from A and B, it will be perpendicular to the line AB at its middle point. 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 will be equal in all their parts. Let the right-angled tri- A angles ABC and DEF have the hypothenuse AC equal to DF, and the side AB B equal to DE: then will the triangles be equal in all their parts. D G E F the triangles will be If the side BC is equal to EF, equal, in accordance with Proposition X. that BC and EF EF are unequal, and that longer. On BC lay off BG equal to EF and draw AG. The triangles ABG and DEF have AB equal to DE, by hypothesis, BG equal to EF, by construction, and 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 line, they will be parallel. Let the two lines AC, BD, be perpendicular to AB: from the same point to the same 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. E 4o. 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. PROPOSITION XIX. THEOREM. 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 will be parallel. Let the lines KC and HD 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. |