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. Q. D_ 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. Produce AB till BC be equal to it, erect (I. 5. cor.) the perpendicular BD equal to AB or BC, bisect I H BC in E (I. 7.), join ED and make Al- - e JE" IB E C EF equal to it; the square of the segment BF is equivalent to the rectangle contained by the whole line BA and its remaining segment AF. For on BC construct the square BG (I. 35.), make BH equal to BF, and draw IHK and FI parallel to AC and BD (I. 23.). Since AB is equal to BD, and BF to BH 3 the remainder AF is equal to HD : and it is farther evident, that FH is a square, and that IC and DK are rectangles. But BC being bisected in E and produced to F, the rectangle under CF, FB, or the rectangle IC, together with the square of BE, is equivalent to the square of EF or of DE (II. 17. cor. 2.). But the square of DE is equivalent to the squares of DB and BE (II. 10.); whence the rectangle IC, with the square of BE, is equivalent to the squares of DB and BE; or, omitting the common square of BE, the rectangle IC is equivalent to the square of DB. Take away from both the rectangle BK, and there remains the square BI, or the square of BF, equivalent to the rectangle HG, or the rectangle contained by BA and AF. Cor. 1. Since the rectangle under CF and FB is equivalent to the square of BC, it is evident that the line CF is likewise divided at B in a manner similar to the original line AB. But this line CF is made up, by joining the whole line AB, now become only the larger portion, to its greater segment BF, which next forms the smaller portion in the new compound. Hence this division of a line being once obtained, a series of other lines all possessing the same property may readily be found, by repeated additions. Thus, let AB be so cut, that the square of BC is equivalent to the rectangle BA, AC: Make successively, BD equal to BA, DE equal to DC, EF equal to EB, and FG equal to FD; the lines CD, BE, DF, and EG, beginning in succession at the points C, B, D, and E, are divided at the points B, D, E, and F, such that, in each of them, the square of the larger part is equivalent to the rectangle contained by the whole and the smaller part.—It is obvious, that this procedure might likewise be reversed. If FD, EB, and DC be made successively equal to FG, EF and DE, the lines DF, BE, and CD will be divided in the same manner at the points E, D and B. Cor. 2. Hence also the construction of another problem of the same nature; in which it is required to produce a straight line AB, such that the rectangle contained by the whole line thus produced and the part produced, shall be equivalent to the square of the line AB itself. Divide AB, by this proposition, in C, so that the rectangle BA, AC is equivalent to the A. Q_3_P square of BC, and produce AB until BD be equal to BC; Then, from what has been demonstrated, it follows that the rectangle under AD and DB is equivalent to $he square of the whole line AB. It will be convenient, for the sake of conciseness, to designate in future this remarkable division of a line, where the rectangle under the whole and one part is equivalent to the square of the other, by the term Medial Section. PROP. XX. THEOR. The square of the side of an isosceles triangle is greater or less than the square of a straight line drawn from the vertex to the base or its extension, by the rectangle contained under its internal or external segments. 1. If BD be drawn from the vertex of the isosceles triangle ABC to a point D in the base; the square of AB exceeds the square of BD, by the rectangle under the segments AD, DC. For (I. 7.) bisect the base AC in E, and join BE. Because the triangles ABE and CBE have the sides AB, AE equal to BC, CE, and the side BE common, they are equal (l. 2.), and consequently the corresponding angles BEA, BEC are equal, and each of them (Def. 4.) a right angle. Wherefore the square of AB is equivalent to the squares of AE and BE (II. 10.); and since AC is cut equally in E and unequally in D, the square of AE is equivalent to the square of DE, together with the rectangles AD, DC (II. 17. cor. 1.); and consequently the square of AB is equivalent to the squares of BE and DE, together with the rectangle AD, DC. Butthe square of BD is equivalent to the squares of BE and DE (II. 10.); whence the square of AB is equivalent to the square of BD, together with the rectangle AD, DC. 2. But the square of the straight line BD drawn from the vertex to any point in the base produced, is greater 1. than the square of AB by the rectangle contained under AD and DC, the external segments of the base. For draw BE, as before, to bisect the base AC. The square of DE is equivalent to the square of AE, together with the rectangle AD, DC, (II. 17. cor. 2.); to 15 *—E-3 each of these, add the square of BE, and the squares of DE and BE,--that is, the square of BD (II. 10.)—are equal to the squares of AE and BE, or the square of BA, together with the rectangle AD, DC. PROP. XXI. THEOR. The difference between the squares of the sides of a triangle, is equivalent to twice the rectangle contained by the base and the distance of its middle point from the perpendicular. Let the side AB of the triangle ABC be greater than BC; and, having let fall the perpendicular BE, and bisected AC in D, the excess of the square of AB above that of BC is equivalent to twice the rectangle contained by AC and DE. For the square of AB is equivalent to the squares of AE and BE (II. 10.), and the square of BC is equivalent to the squares CE and BE; wherefore the excess of the square of AB above that of BC is equivalent H to the excess of the square of AE above, s that of CE. But the excess of the square of AE above that of CE, is (II. 17.) equivalent to the rectangle contained by their sum AC and their difference, which is evi - F |