PROP. XXXI. PROB. To draw a straight line through a given point parallel to a given straight line. Let A be the given point; and BC the given straight line; it is required to draw a straight line through the point A, а make the angle DAE equal to the angle ADC; and produce the straight line EA to F. Because the straight line AD, which meets the two straight lines BC, EF, F makes the alternate angles EAD, ADC equal to one another, EF is parallel to BC. Therefore the straight line EAF is drawn through the given point A parallel to the given straight line BC. Which was to be done. [In making the angle DAE equal to the angle ADC, use the method pointed out in the Note to Proposition XXIII. Thus: BOOK I. a xxiii. 1. b xxvii. 1. BOOK I. * xxxi. 1. PROP. XXXII. THEOR. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles. And the three interior angles of every triangle are equal to two right angles. Let ABC be a triangle, and let one of its sides BC be produced to D; (1.) the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; (2.) and the three interior angles of the triangle, viz. ABC, BCA, CAB, are together equal to two right angles. Through the point C draw CE parallela to the straight line AB, B (1.) And because AB is parallel to CE, and AC meets them, xxix. 1. the alternate angles BAC, ACE are equal.b Again, because AB is parallel to CE, and BD falls upon them, the exterior angle ECD is equal to the interior and opposite angle ABC; but the angle ACE was shown to be equal to BAC; and the angles ACD, ACB are equal to the three angles CBA, BAC, ACB; E D xviii. i. but the angles ACD, ACB are equal to two right angles; therefore also the angles CBA, BAC, ACB are equal to two right angles. Wherefore if a side of a triangle, &c. Q. E. D. COR. 1. All the interior angles of any rectilinear figure, BOOK I. together with four right angles, are equal to twice as many right angles as the figure has sides. For any rectilinear figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. And, by the preceding Proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure; and the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles; that is, together with four right angles. B [The triangle furnishes an instance of this, for all its angles are together equal to two right angles; or all its angles, together with four right angles, are equal to six right angles, or twice as many right angles as the figure has sides. A square (see Def. 30.) and an oblong (see Def. 31.) are other obvious instances of this, for they have four right angles, and thus all their angles, together with four right angles, are together equal to eight right angles, that is, twice as many right angles as the figure has sides. The same may be exhibited in any four-sided figure by joining two of its opposite angular points. Thus : The interior angles of ABDC are together equal to four right angles, for they are made up of all the angles of the two triangles into which the figure C B is divided, that is, twice two right angles.] D • 2 Cor. xv. 1. BOOK I. b xiii. i. COR. 2. All the exterior angles of any rectilinear figure are together equal to four right angles. is equal to two right angles; with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure; that is, by the foregoing corollary, A they are equal to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal to four right angles. [This also may be advantageously illustrated in the triangle and the square.] PROP. XXXIII. THEOR. The straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal (1.) and (2.) parallel. Let AB, CD be equal and parallel straight lines, and joined towards the same parts by the straight lines AC, BD, AC, BD are also equal and parallel. Join BC; (1.) And because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal"; and because AB is equal to CD, and BC common to the two triangles, ABC, DCB, the two sides AB, BC are equal to the two DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal to the base BD, B D (2.) And the triangle ABC to the triangle BCD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD; and it was shown to be equal to it. Therefore, straight lines, &c. Q. E. D. BOOK I. a xxix. 1. b iv. 1. © xxvii. 1. |