From the extremity B draw, by the last proposition, BC perpendicular to BA and equal to it, and, from the points A and C ID C with the distance BA or BC describe T two circles intersecting each other in the point D, join AD and CD; the quadrilateral figure ABCD is the square required. . For, by this construction, the figure has all its sides equal, and one of its angles ABC a right angle; which comprehends the whole of the definition of a square. PROP. XXXVI. PROB. To divide a given straightline into any number of equal parts. Let it be required to divide the straight line AB into a given number of equal parts, suppose five. From the point A and at any oblique angle with AB, draw a straight line AC, in which take the portion AD, and repeat it five times from A to C, join CB, and from the several points of section D, E, F, and G draw the parallels DH, EI, FK, and GL, (I. 23.), cutting AB in H, I, K, and L: AB is divided at these points into five equal parts. For (I. 23.) draw DM, EN, FO, and GP parallel to AB. And because DH is parallel to EM, the exterior angle ADH is equal to DEM (I. 22.); and, for the same reason, since AH is parallel to DM, the angle DAH is equal to EDM. Wherefore the triangles ADH and DEM, having two angles respectively equal and the interjacent sides AD, DE—are (I.20.) equal, and consequently AH is equal to DM. In the same manner, the triangle ADH is proved to be equal to EFN, to FGO, and GCP; and therefore their bases, EN, FO, and GP are all equal to AH. But these lines are equal to HI, IK, KL, and LB, for the opposite sides of parallelograms are equal (I. 26.). Wherefore the several segments AH, HI, IK, KL, and LB, into which the straight line AB is divided, are all equal to each other. *. Scholium. The construction of this problem may be facilitated in practice, by drawing from B in the opposite direction a straight line parallel to AC, and repeating on both of them portions equal to the assumed segment AD, but only four times, or one fewer than the number of divisions required ; then joining D, the first section of AC, with the last of its parallel, E with the next, and so on till G, which connecting lines are (I. 27.) all parallel, and consequently the former demonstration still holds. EE,TEMENTS OF GEOMETRY. BOOK II. DEFINITIONS. 1. IN a right-angled triangle, the side that subtends the right angle is termed the hypotenuse; either of the sides which contain it, the base ; and the other side, the perpendicular. 2. The altitude of a triangle is a perpendicular let fall from the \ - 27 vertex upon the base or its extension. 3. The altitude of a trapezoid is the perpendicular drawn from one of its | |\, parallel sides to the other. 4. The complements of rhomboids about the diagonal of a rhomboid, are the spaces required to complete the rhomboid; and the defect AE/ of each rhomboid from the whole figure, /~/ is termed a gnomon. 5. A rhomboid or rectangle is said to be contained by any two adjacent sides. A rhomboid is often indicated merely by the two letters placed at opposite corners. PROP. I. THEOR. Triangles which have the same altitude, and stand on the same base, are equivalent. The triangles ABC and ADC which stand on the same base AC and have the same altitude, contain equal spaces. For join the vertices B, D by a straight line, which produce both ways; and from A draw AE (I. 23.) parallel to CB, and from C draw CF parallel to AD. Because the triangles ABC, ADC have the same altitude, the straight line EF is parallel to AC (I. 24.), and consequently the figures CE and AF T} D F' are parallelograms. Wherefore EB, being equal to AC (I. 26.) which is equal to DF, is itself equal to DF. Add BD to each, and ED is equal to BF; but EA is equal to BC (I. 26.), and the interior angle AED is equal to the exterior angle CBF (I. 22.). Thus the two triangles EDA, BFC have the sides ED, EA equal to BF, BC, and the contained angle AED equal to CBF, and are therefore equal (I. 3.). Take these equal triangles CBF and EDA from the whole quadrilateral space AEFC, and there remains the rhomboid AEBC equivalent to ADFC. Whence the triangles ABC and ADC, which are the halves of these rhomboids (I. 26. cor.), are likewise equivalent. Cor. Hence rhomboids on the same base and between the same parallels, are equivalent. PROP. II. THEOR. Triangles which have the same altitude, and stand on equal bases, are equivalent. The triangles ABC, DEF, standing on equal bases AC and DF and having the same altitude, contain equal spaces. For let the bases AC, DF be placed in the same straight line, join BE, and produce it both ways, draw AG and DH parallel to CB and FE (I. 23.), and join AH, CE. Because the triangles ABC, DEF are of equal altitude, GE is parallel to AF (I. 24.), and GC, HF are parallelograms. But AC, being equal to G DF, and DF equal (I. 26.) to IB II E. HE, must also be equal to HE, and therefore (I. 27.) AE is C ID JT a rhomboid or parallelogram. Whence the rhomboid GC is equivalent to AE(II. 1. cor.), and this again is, for the same reason, equivalent to HF : . consequently GC is equivalent to HF, and therefore their halves or (I. 26. cor.) the triangles ABC and DEF are equivalent. Cor. 1. Hence rhomboids on equal bases and between the same parallels, are equivalent. Cor. 2. Hence triangles which have the same vertex, and equal bases in the extension of the same straight line, are equivalent; and hence straight, lines drawn from the vertex of a triangle to equal sections of the base, will likewise divide it into equivalent triangles. |