OF PERPENDICULAR AND OBLIQUE LINES. THEOREM IX. If two straight lines have two points common they will coincide throughout their whole extent. If A and B are points common to two straight lines, they will coincide between A and B (A. XIII.). A B F E D If possible, we will suppose that when produced, they begin to separate at C, the one taking the direction CD, and the other the direction CE. Draw CF perpendicular to AC; then since the lines ACE and ACD are each straight, we have the angle FCD and FCE equal, each being equal to a right angle (T. I., C. I.); that is, the whole is equal to one of its parts, which is impossible (A. VIII.). It is therefore absurd to suppose these lines can separate when produced. Hence, if two straight lines have two points common they will coincide throughout their whole extent. THEOREM X. Through a given point in a straight line only one perpendicular can be drawn to this line. A D E C B If there could be two perpendiculars, as CD and CE, the angles BCD and BCE would be equal, each being a right angle; that is, the whole would be equal to its part, which is impossible (A. VIII.). Hence, it is absurd to suppose that more than one perpendicular can be drawn to a given line through any one of its points. THEOREM XI. From a given point without a straight line, only one perpendicular can be drawn. Let C be the point, and AB the given line. If possible, suppose we can draw the two perpendiculars CD and CE. Revolve C the figure CDE about DE as a hinge, until it returns into its primitive plane on the opposite side of AB, having the position FDE. Since each of the angles FDE and CDE is right, CDF is a straight line (T. I., C. IV.). By the same reasoning we have CEF a straight line. That is, we have two straight lines joining the points C and F, which is impossible (A. XIII.). Hence, from a point without a straight line only one perpendicular can be drawn to this line. A B D E F THEOREM XII. If from a point without a line, a perpendicular be drawn, and several oblique lines : I. The perpendicular will be shorter than any oblique line. II. Any two oblique lines which terminate at equal distances from the foot of the perpendicular, will be equal. III. Of two oblique lines terminating at unequal distances from the foot of the perpendicular, the one at the greater distance will be the longer. Let A be the given point, BC the given line, AD perpendicular, and AB, AE, and AC oblique lines. A E .C G B F First. As in the last Theorem, suppose ADC to revolve about BC into the position FDC, on the opposite side of BC in its primitive plane. We have FD=AD; FE=AE; and FC=AC. But since each of the angles FDE and ADE is a right angle, ADF is a straight line (T. I., C. IV.), and therefore shorter than the broken line AE+EF; hence, AD, the half of AF, is less than AE, the half of AE+EF. That is, the perpendicular is shorter than any oblique line. Secondly. If we suppose the figure ADB to revolve about AD, the point B will coincide with E, since DB is equal to DE, and the angle ADB is equal to ADE, each being a right angle. Hence, the oblique line AB will coincide with AE. That is, two oblique lines which terminate at equal distances from the foot of the perpendicular are equal. Thirdly. Returning to the figure as first revolved, we have, since E is a point within the triangle ACF, AC + CF>AE+EF (T. VIII.); hence, AC, the half of AC+CF, is greater than AE, the half of AE+EF. That is, the oblique line terminating farther from the foot of the perpendicular is the longer. Cor. I. The perpendicular measures the shortest distance from a point to a line. Cor. II. From the same point without a straight line, only two equal oblique lines can be drawn, one on each side of the perpendicular. THEOREM XIII. If through the middle point of a straight line, a perpendicular be drawn: I. Any point in this perpendicular will be equally distant from the extremities of this line. II. Any point without the perpendicular will be unequally distant from, the extremities of this line. Let AB be the given line, DE a perpendicular through C, its middle point. A F G D E طح B C First. Let D be any point in this perpendicular. Drawing DA and DB, we know these lines to be equal, since they terminate at equal distances from C, the foot of the perpendicular DC (T. XII.). That is, D is equidistant from A and B. In the same manner we may show that any other point of this perpendicular, as E or F, is equidistant from the extremities of the given line. Secondly. Suppose G to be a point without the perpendicular. If we draw GA and GB, one of these lines must cut the perpendicular. We will suppose GA to cut it at D. Draw DB, then GB will be less than GD+DB (T. VII.). But, we already have DB = DA, hence GB is less than GD +DA, or less than GA. That is, the point G is unequally distant from the extremities of the given line. In the same manner we can show that any other point without the perpendicular is unequally distant from the extremities of this line. Scholium. The most distant extremity A of this line will be on the side of the perpendicular opposite the point G. Cor. I. Any point equidistant from the extremities of a straight line is situated in the perpendicular which bisects this line. Cor. II. If a straight line have any two of its points equidistant from the extremities of a second line, it will be perpendicular to the second line and bisect it. THEOREM XIV. If a line be drawn bisecting a given angle, that is, dividing it into two equal angles : I. Any point in this bisecting line will be equidistant from the sides of the angle. II. Any point without this bisecting line will be unequally distant from the sides of the angle. First. Let the given angle BAC be bisected by the line AD, then will any point as D in this line be equidistant from the lines AB and AC. The distance of D from these lines will be measured by the perpendiculars DE and DF (T. XII., C. I.). B H A E F D C If we conceive the figure folded about the line AD, so that the portion on the left of AD may be superposed upon the portion on the right, the line AB will then coincide with AC, since the angle BAD = CAD. Consequently the perpendiculars DE and DF will coincide (T. XI.), and are therefore equal. Secondly. Suppose G to be a point without the bisecting line. If we draw the perpendiculars GF and GH, one of these must cut the bisecting line. Let GF cut it at D. Draw DE perpendicular to AB, and join G and E: then will GH be less than GE (T. XII.). But GE is less than GD + DE (T. VII.); therefore, GH is less than GD + DE, or less than GD + DF, since by the first case of this Theorem DF=DE. Hence, finally, GH is less than GF. Cor. I. Any point equidistant from the sides of a given angle is situated on the line bisecting this angle. Cor. II. If a straight line have two of its points equidistant from the sides of a given angle, it will bisect this angle. OF PARALLEL LINES. THEOREM XV. When two lines are perpendicular to the same line, they are parallel. If the lines AB and CD are perpendicular to GH they will be parallel. G E A B For, since they are perpendicular to GH, we have the angles BEG and DFE equal, c each being a right angle; hence, these lines have the same direction, and are therefore parallel (D. X.). D F H Cor. A line which is perpendicular to one of two parallels, is also perpendicular to the other. THEOREM XVI. Two parallel lines are throughout their whole extent equally distant. FLH A B C D EK G Suppose AB and CD to be parallel. Through any two points, as E and G, of the line CD, draw EF and GH perpendicular to CD, they will also be perpendicular to AB (T. XV.), and they will measure the distance of E and G respectively from the line AB (T. XII., C. I.). We are now to prove these lines equal. Through K, the middle of EG, draw KL perpendicular to CD, and it will also be perpendicular to AB. Now, suppose the portion of the figure on the left of KL to revolve about KL as a hinge, until it returns into its primitive plane, on the right of KL. The angles at K and L being right, KE will take the direction of KG, and LF the direction of LH, and E will coincide with G, since KEKG, and since EF and GH are each perpendicular to CD, they must coincide, which proves them equal. Cor. Two parallels cannot meet, however far they are pro duced. |