For through B draw BI (1.23.) parallel to AD, make AF equal to AB, and through F draw FH parallel to DE. It is manifest that the spaces AG, GE, DG and CG, into which the square of AC is divid- T ed, are all rhomboidal and rectangu- * Er lar. And because AB is equal to F AF, and the opposite sides equal, the figure AG is equilateral, and having a right angle at A, is hence a A #-c square. Again, AD being equal to AC, take away the equals AF and AB, and there remains DF equal to BC, and consequently IG equal to GH (I. 26.): wherefore IH is likewise a square. The rectangle DG is contained by the sides FG and DF, which are equal to AB and BC; and the rectangle CG is contained by the sides GB and GH, which are likewise equal to AB and BC. Consequently the whole square of AC is composed of the two squares of AB and BC, together with twice the rectangle contained by these lines. PROP. XVI. THEOR. The square described on the difference of two straight lines, is equivalent to the squares of those . lines, diminished by twice their rectangle. Let AC be the difference of two straight lines AB and BC; the square of AC is equivalent to the excess of the two squares of AB and BC above twice their rectangle. For let the squares of AB, BC and AC be completed, and produce CE and DE the sides of the latter to Hand I. It is evident, that GE is equal to BL or the square of BC; to each add the intermediate rect- | angle EB, and GC is equal to IL; R H. —g but the rectangle under AB and BC is equal to the rectangle IL, ID —H: s which is also equal to DG. From the compound surface CAFGBKL, or the squares of AB and BC, take Å. C B away the space DFGBKLC, or the rectangles IL and DG, that is, - K twice the rectangle under AB and BC –and there remains ADEC, or the square of the difference AC of the two lines The rectangle contained by the sum and difference of two straight lines, is equivalent to the difference of their squares. Let AB and BD be two continuous straight lines, of which AD is the sum and AC the difference; the rectangle under AD and AC, is equivalent to the excess of the square of AB above that of BC. For, having made AG equal to AC, draw GH parallel to AD (I. 23.), and CI, DH parallel to AE. Because GK is equal to KC or HD, and EG is equal to CB or BD, the rectangle EK is equal to LD (II. 2, cor.); and # I To consequently, adding the rectangle , BG to each, the space AEIKLB G E-TETH is equivalent to the rectangle AH. But this space AEIKLB is the ex- A. CTBTI) cess of the square of AB above IL or the square of BC; and the rectangle AH is contained by AD and DH or AC. Wherefore the rectangle under AD and AC is equivalent to the difference of the squares of AB and BC. Cor. 1. Hence if a straight line AB be bisected in C and cut unequally in D, the rectangle under the unequal segments AD, DB, together with the square of CD, the interval between the points of section, is equivalent to the square of AC, the half line. For AD is the sum of AC, CD, and DB is A. Q. P R evidently their difference; whence, by the Proposition, the rectangle AD, DB is equivalent to the excess of the square of AC above that of CD, and consequently the rectangle AD, DB, with the square of CD, is equal to the square of AC. Cor. 2. If a straight line AB be bisected in C and produced to D, the rectangle contained by AD the whole line 'thus produced, and the produced part DB, together with the square of the half line AC, is equivalent to the square of CD, which is made up of the half line and the produced part. For AD is the sum of AC, CD, and DB is their difference; whence A. Q. P. D. the rectangle AD, DB is equivalent to the excess of the square of CD above AC, or the rectangle AD, DB, with the square of AC, is equivalent to the square of CD. Scholium. If we consider the distances DA, DB of the point D from the extremities of AB as segments of this line, whether formed by internal or external section; both corollaries may be comprehended under the same enunciation, That, if a straight line be divided equally and unequally, the rectangle contained by the unequal segments is equivalent to the difference of the squares of the half line and of the interval between the points of section. PROP. XVIII. THEOR. The sum of the squares of two straight lines, is equivalent to twice the squares of half their sum and of half their difference. Let AB, BC be two continuous straight lines, D the middle point of AC, and consequently AD half the sum of these lines and DB half their difference; the squares of AB and BC are together equivalent to twice the square of AD, with twice the square of DB. For erect (I. 5. cor.) the perpendicular DE equal to AD or DC, join AE and EC, through B and F draw (I. 23.) BF and FG parallel to DE and AC, and join AF. Because AD is equal to DE, the angle DAE (I. 10.) is equal to DEA, and since (I. 30. cor.) they make up together one right angle, each of them must be half a right angle. In the same manner, the angles DEC and DCE of the triangle EDC are proved to be each half a right angle; consequently the angle AEC, composed of AED and CED, is equal to a whole right angle. And in the triangle FBC, the angle CBF being equal to CDE (I. 22.) which is a right . E angle, and the angle BCF being half F a right angle—the remaining an -A- TD B C gle BFC is also half a right angle (I. 30.), and therefore equal to the angle BCF; whence (I, 11.) the side BF is equal to BC. By the same reasoning, it may be shown, that the right angled triangle GEF is likewise isosceles. The square of the hypotenuse EF, which is equivalent to the squares of EG and GF (II. 10.), is therefore equivalent to twice the square of GF or of DB; and the square of AE, in the right angled triangle ADE, is equivalent to the squares of AD and DE, or twice the square of AD. But since ABF is a right angle, the square of AF is equivalent to the squares of AB and BF, or BC; and because AEF is also a right angle, the square of the same line AF is equivalent to the squares of AE and EF, that is, to twice the squares of AD and DB. Wherefore the squares of AB, BC are together equivalent to twice the squares of AD and DB. Cor. Hence if a straight line AB be bisected in C and cut unequally in D, whether by internal or external section, the squares A CD B of the unequal segments AD and DB + C B P are together equivalent to twice the square of the half line AC, and twice the square of CD the interval between the points of division. PROP. XIX. PROB. To cut a given straight line, such that the square of one part shall be equivalent to the rectangle contained by the whole line and the remaining part. Let AB be the straight line which it is required to divide into two segments, such that the square of the one shall be equivalent to the rectangle contained by the whole line and the other. |