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

A

C D B

is the sum of AC, CD, and DB is 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 equivalent 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 extended line, and its produced part DB, together with the square of the half line AC, is equivalent to the square of CD, which is made up of this half line and of the part produced. For AD is the sum of AC, CD, and DB is their difference; whence 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.

B D

Scholium. If, enlarging our views, 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 angle, and the angle BCF being half a right angle-the remaining angle 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. Wherefore, the square of the hy

E

A.

D B

potenuse EF, which is equivalent to the squares of EG and GF (II. 10.), is 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 likewise 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 of DB. Consequently 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 in ternal or external section, the squares of the unequal segments AD and DB are together equivalent to twice the

CD B

A

C

B D

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, BF and AF, such that the square of the one shall be equivalent to the rectangle contained by the whole line and the other.

D

༢ ༢

K

Produce AB till BC be equal to it, erect (I. 5. cor.) the perpendicular BD equal to AB or BC, bisect BC in E (I. 7.), join ED and make 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.

AF

B E C

For complete 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; 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 the hypotenuse 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 the 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 peculiar division of any 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 AC B D

Ε

F

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. Even if the section of AB were assumed at first inexact, the series of combinations would always approach to greater accuracy.

The 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. For, by this proposition, divide AB in C, so that the rectangle BA, AC shall be AC

АС B

D

equivalent to the 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 must be equivalent to the 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.