BOOK I. RECTILINEAR FIGURES. PERPENDICULAR AND OBLIQUE LINES. Proposition 1. Theorem. 48. All straight angles are equal to one another.* Hyp. Let AB, AC be the arms of a st. whose vertex is A, and DE, DF the arms of another st. whose vertex is D. B + A E F D To prove < BAC = EDF. Proof. Because the BAC is a st., .. BA and AC are in the same st. line BAC. Because the EDF is a st. 4, .. ED and DF are in the same st. line EDF. (21) (21) Now if the BAC be applied to the EDF so that the vertex A shall coincide with the vertex D and the arm AB with the arm DE, then the arm AC will coincide with the arm DF, because, if two st. lines have two points in common they will coincide throughout their whole length, and form but one st. line (Ax. 11). ... Z BAC = ZEDF. Q. E. D. 49. COR. 1. All right angles are equal to one another. † (21), also (Ax. 7). 50. COR. 2. The complements of equal angles are equal to each other, and the supplements of equal angles are equal to each other. 51. COR. 3. At a given point in a given straight line only one perpendicular can be drawn to that line. * The Elements of Plane Geometry, by Association for the Improvement of Geometric Teaching, p. 20a. This is Euclid's 11th axiom. 14 Proposition 2. Theorem. 52. If one straight line meet another straight line, the sum of the two adjacent angles is equal to two right angles. Hyp. Let the st. line DC meet the .'. Z ACD + ≤ DCB = ZACE + E D -B ECD + ≤ DCB. If equals be added to equals, the sums will be equal (Ax. 2). ...Z ACE + ECB = ACE + ECD + 2 DCB. Hence If equals be added to equals, the sums will be equal (Ax. 2). SCH. The angles ACD, DCB are supplementary adjacent angles (25). 53. COR. If one of the angles ACD, DCB is a right angle, the other is also a right angle. The sum of all the angles on the same side of a straight line, at a common point, is equal to two right angles; for their sum is equal to the sum of the two adjacent angles ACD, DCB. * The student should quote every reference in full. Proposition 3. Theorem. 54. Conversely, if the sum of two adjacent angles is equal to two right angles, their exterior sides are in one straight line. Hyp. Let /ACD + / DCB = 2 rt. s. То prove that AC and CB are in one st. line. Proof. If CB be not in the same st. line as AC, let CE be in the same st. line as AC. Then ACD + DCE = 2 rt. s. But being sup. adj. ≤8 (52). ACD + ≤ DCB = 2 rt. s. (Hyp.) ... ACD + DCE = ACD + 2 DCB. (Ax. 1) ≤ 2 Take away the common / ACD. .. / DCE / DCB, = (Ax. 3) which is impossible (Ax. 8), unless CE coincides with CB. ... AC and CB are in one st. line. Q.E.D. EXERCISES. 1. Find the number of degrees in an angle if it is three times its complement. 2. Find the number of degrees in an angle if its complement and supplement are together equal to 110°. Proposition 4. Theorem. 55. If two straight lines intersect each other, the vertical angles are equal. Hyp. Let the st. lines AB and CD cut at E. To prove Proof. < AEC BED. CEA + ≤ CEB = 2 rt. ≤s, being sup. adj. ≤8 (52). AEC+ AED = 2 rt. LS, being sup. adj. 48 (52). :: 2 CEA + ▲ CEB = ▲ AEC + AED. All rt. 8 are equal to one another (49). 56. COR. 1. If two straight lines intersect each other, the four angles which they make at the point of intersection, are together equal to four right angles. If one of the four angles is a right angle, the other three are right angles, and the lines are mutually perpendicular to each other. 57. COR. 2. If any number of straight lines meet at a point, the sum of all the angles made by consecutive lines, in a plane, is equal to four right angles, Proposition 5. Theorem. 58. From a point without a straight line only one perpendicular can be drawn to that line; and this perpendicular is the shortest distance from the point to the line. Hyp. Let AB be the given st. line, P the point without the line, PC a from P to AB, and PD other line from P to AB. (1) To prove that PD is not 1 to AB. any Proof. Let the part of the plane above the line AB, and which contains P, be revolved about AB as an axis until the point P comes into the position P'. A P с D Then, since ACP is a rt. 2, Z ACP' is also a rt. . (29) If ADP is a rt. Z, ADP' is also a rt. Z. .. PCP' is a st. line. .. PDP' is a st. line; (54) (29) (54) that is, between two points there are two straight lines, which is impossible. (Ax. 11) .. ADP is not a rt. ▲, and .. PD is not 1 to AB. (20) .. PC is the only 1 to AB from the point P. Proof. Since in the revolution of CPD about AB as an axis, the point P comes into the position P', 59. SCH. By the distance of a point from a line is meant the shortest distance, that is, the length of the perpendicular from the point to the line. |