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likewise equal and parallel to each other, and that the figure AKLC is a parallelogram or rhomboid.—This rhomboid is equivalent to the two rhomboids BD and BF. For produce HB to meet the base AC in I. And because the rhomboids KI and AH stand on the same base AK and between the same parallels, they are equivalent (II. 1. cor.); but the rhomboids AH and BD, standing on the same base AB and between the same parallels, are also equivalent. Whence KI is equivalent to BD. And in the same manner, it may be proved that LI is equivalent to BF. Consequently the whole rhomboid KC is equivalent to the two rhomboids BD and BF. If the triangle ABC be right-angled at B, this theorem will pass into a case of the twenty-sixth of Book VI. ; the rhomboid, described on the hypotenuse, being equivalent to the similar rhomboids described on the two sides. When these rhomboids become squares, the proposition becomes the same as the tenth; the only difference in the construction being, that a square AKOC (p. 52) is constructed above the hypotenuse AC, instead of the square ADEC constructed below it.
8. From the proposition in the last article, an important theorem may be derived, which deserves a place in an elementary work: In any triangle, the square described on the base is equivalent to the rectangles contained by the two sides and their segments intercepted from the base by perpendiculars let fall upon them from its opposite extremities. Let the perpendiculars AP, CN be let fall from the points A, C upon the opposite sides BC and AB of the triangle ABC; the square of AC is equivalent to the rectangles contained by AB, AN, and by BC, CP. For complete the rhomboids ADHB and CFHB, and let fall the perpendiculars BR and BS upon DH and FH.
It is manifest, that the rhomboids AH and CH are equivalent to the square of AC. But the rhomboid AH is equivalent to the rectangle contained by AB and BR II, 1.cor.). Comparing the triangles BHR Fl and ACN ; the angle BRH, being a right angle, is equal to ANC; and the It two acute angles BHR and RBH, being D together equal to a right angle, are e- Noll qual to DAN and NAC; but DAB is TN/ equal to DHB (I.26.), whence the angle RBH is equal to NAC. These triangles BHR and ACN, having thus two angles X *: respectively equal, and the corresponding side BH in the one equal to AD or AC in the other, are therefore equal p F
(I. 20.), and consequently the side BR \
is equal to AN. The rectangle AB and BR, which is equivalent to the rhom- P boid AH, is hence equivalent to the rectangle contained by AB and AN (II. A. C 1. cor.).
In the same manner, it may be demonstrated, by comparing the triangles BHS and PAC, that the rectangle under BC and BS, which is equivalent to the rhomboid CH, is equivalent. to the rectangle contained by BC and CP. Wherefore the two rectangles of AB, AN and BC, CP are together equivalent to the square described on AC.
If the triangle ABC be right-angled at the vertex B, the perpendiculars CN and AP will evidently meet at the vertex, and consequently the rectangles AB, AN and BC, CP will become the squares of AB and BC. And hence the beautiful Proposition II, 10, is derived, being only a remarkable case of a much more general property.
9. Proposition tenth. It may be proper to notice likewise an extension of this beautiful proposition, which is easily demonstrated, after a similar mode, from the decomposition of the figure.
Equilateral triangles described on the sides of a right-angled triangle, are together equivalent to an equilateral triangle described on the hypotenuse. Let ABC be a right-angled triangle, around which are constructed the equilateral triangles ADB, BEC and CFA ; the triangles ADB and BEC are equivalent to CFA. For let fall the perpendiculars DG, BH and FI, and join CD, BF, CG, BI and HF. It is evident (I. 21.) that the perpendiculars DG and FI bisect the bases AB and AC, and divide the triangles ADB and CFA into two equal triangles. But the angle DAB is equal to CAF, being angles of an equilateral triangle : . add BAC to each, and the whole angle DAC is equal to BAF. But the containing sides DA and AC are respectively equal to BA and AF, and consequently (I. 3.) the triangle ADC is equal to ABF. Now the triangle ADC is composed of the three triangles ACG, ADG, and DCG, and the triangle ABF is composed of ABI, AFI, and FBI; but, since AB and AC are bisected in G and H, the triangles ACG and ABI are (II. 2.) halves of the original triangle ABC, and consequently equivalent to each other. Wherefore the remaining triangles ADG and DCG are together equivalent to AFI and FBI. But DG and CB being both perpendicular to AB, are (I.22.) parallel ; and, for the same reason, BH is parallel to FI. Whence (II. 1.) the triangle DCG is equivalent to DBG, and the triangle FBI equivalent to FHI; and therefore the triangles ADG and DBG, or the whole triangle ADB, must be equivalent to AFI and FHI, or the whole triangle AFH.—In like manner, it may be shown that the triangle BEC is equivalent to the triangle CFH ; and consequently the equilateral triangles ADB and BEC are equivalent to AFH and CFH, which make up the whole triangle AFC. This demonstration is the second of those given by the celebrated Italian geometer Torricelli, the favourite disciple of Galileo, and inventor of the barometer.
10. A useful proposition may be introduced here : The square described on a straight line, is equivalent to the squares of the segments into which it is divided, and twice the rectangles contained by each pair of these segments. The square of AB is equivalent to the squares of AC, of CD and of DB, with twice the rectangles of AC, CD, of AC, DB, and of CD, DB. For make AE and EF equal to AC and CD draw EM, FL parallel to AB, and CH, DI parallel to AG. It is manifest that AO is the square of AC, OQ the square of CD, and QK G H H S the square of DB. Nor is it less obvio F. P ous that the two rectangles CN and EP are contained by AC, CD, that the two rectangles NL and PI are contained by CD, DB, and that the two rectangles DM and FH are contained by AC, DB. But those squares and those double rectangles complete the whole square of AB. Wherefore the truth of the Proposition is established. Cor. Hence if a straight line be divided into three portions, the squares of the double segments AD, BC, together with twice the rectangle under the extreme segments AC, BD, are equivalent to the squares of the whole line AB and of the intermediate segment CD. For the squares FD, HM, together with the equal rectangles GP, NB, evidently fill up the whole square AB, with the repetition of the internal square OQ ; that is, the squares of AD and BC, with twice the rectangle AC, DB, are equivalent to the squares of AB and CD.
11. Since rectangles correspond to numerical products, the properties of the sections of lines are easily derived from symbolical arithmetic or algebra:
7. In Prop. 19. let the whole line be denominated by a, and its greater segment by a then road(a—w), and *-i-ar-a’, 2 whence a = =E #–4 ==Ea(V3–4). Hence, if unit represent the whole line, the greater segment is .61803398428, &c. and the smaller segment .38196601572, &c. From Cor. 1. an extremely neat approximation is likewise obtained. Assuming the segments of the divided line as at first equal, and each denoted by 1, the following successive numbers will result from a continued summation: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, &c, which are thus composed, 1+2=3, 2+3=5, 3+5=8, 5+ 8= 13, 8+ 13=21, &c. These numbers form, therefore, a simple recurring series, a kind of approximation which was first noticed in this actual case early in the seventeenth century, by Gerard, an ingenious Flemish mathematician. Hence, if the original line contained 144 equal parts, its greater segment would include 89, and its smaller segment 55 of these parts, very nearly; but 55 x 144 = 7920, being only one less than 7921, the square of 89.
12. Proposition nineteenth, cor. 2. This problem may, however, be constructed somewhat differently, without employing the collateral properties.