The same process may be applied to every other polygon; for, by successively diminishing the number of its sides, one being retrenched at each step of the process, the equivalent triangle will at length be found. COR. Since a triangle may be converted into an equivalent rectangle it follows that any polygon may be reduced to an equivalent rectangle. PROP. H. PROB. To find the side of a square that shall be equivalent to the sum of two squares Draw the two indefinite lines AB, AC, perpendicular to each other. Take AB equal to the side of one of the given squares, and AC equal to the other; join BC: this will be the side of the square required. For the triangle BAC being right angled, the square constructed upon BC (47. 1.) is equal to the sum of the squares described upon AB and AC. D A B SCHOLIUM. A square may be thus formed that shall be equivalent to the sum of any number of squares; for a similar construction which reduces two of them to one, will reduce three of them to two, and these two to one, and so of others. PROP. I. PROB. To find the side of a square equivalent to the difference of two given squares. Draw, as in the last problem, (see the fig.) the lines AC, AD, at right angles to each other, making AC equal to the side of the less square; then, from C as centre, with a radius equal to the side of the other square, describe an arc cutting AD in D: the square described upon AD will be equivalent to the difference of the squares constructed upon AC and CD. For the triangle DAC is right angled; therefore, the square described upon DC is equivalent to the squares constructed upon AD and AC: hence (Cor. 1. 47. .), AD2-CD2-AC. PROP. K. PROB. A rectangle being given, to construct an equivalent one, having a side of given length. Let AEFH be the given rectangle, and produce one of its sides, as AH, till C G COR. A polygon may be converted into an equivalent rectangle, having one f its sides of a given length. ELEMENTS or GEOMETRY. BOOK II. DEFINITIONS. 1 EVERY right angled parallelogram, or rectangle, is said to be contained by any two of the straight lines which are about one of the right angles. Thus the right angled parallelogram AC is called the rectangle contain"ed by AD and DC, or by AD and AB, &c. For the sake of brevity, "instead of the rectangle contained by AD and DC, we shall simply say "the rectangle AD. ĎC, placing a point between the two sides of the "rectangle." A. In Geometry, the product of two lines means the same thing as their rectangle, and this expression has passed into Arithmetic and Algebra, where it serves to designate the product of two unequal numbers or quantities, the expression square being employed to designate the pro duct of a quantity multiplied by itself. The arithmetical squares of 1, 2, 3, &c. are 1, 4, 9, &c. 2 In every parallelogram, any of the PROP. I. THEOR. If there be two straight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E; the rectangle A.BC is equal to the several rectangles A.BD, A.DE, A.EC. From the point B draw (Prop. 11.1.) BF at right angles to BC, and make BG equal (Prop. 3. 1.) to A; and through G draw (Prop. 31. 1.) GH parallel to BC; and through D, E, C, draw DK, EL, CH parallel to BG; then BH, BK, DL, and EH are rectangles, and BH= BK+DL+EH. But BHBG.BC= A.BC, because BG=A: Also BK = BG.BD=A.BD, because BG=A; and DL=DK.DE= A.DE, because (34. 1.) DK=BG=A. B G F DEC KLH A In like manner, EH=A.EC. Therefore A.BC=A.BD+A.DE+A.EC; that is, the rectangle A.BC is equal to the several rectangles A.BD, A.DE, A.EC. SCHOLIUM. The properties of the sections of lines, demonstrated in this Book, are easily derived from Algebra. In this proposition, for instance, let the seg ments of BC be denoted by b, c, and d; then, A(b+c+d)=Ab+Ac+Ad. If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square of the whole line. Let the straight line AB be divided into any two parts in the point C; the rectangle AB.BC, together with the rectangle AB.AC, is equal to the square of AB; or AB.AC+AB.BC=AB2. On AB describe (Prop. 46. 1.) the square ADEB, and through C draw CF (Prop. 31. 1.) parallel to AD or BE; then AF÷CE=AE. But AF AD.AC=AB.AC, because AD=AB; CE=BE.BC=AB.BC; and AE=AB2. Therefore AB.AC+AB.BC=AB2. SCHOLIUM. A D C B FE This property is evident from Algebra: let AB be denoted by a, and the segments AC, CB, by b and d, respectively; then, a=b+d; therefore, multiplying both members of this equality by a, we shall have a2=ah+ad PROP. III. THEOR. If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part. Let the straight line AB be divided into two parts, in the point C; the rectangle AB.BC is equal to the rectangle AC.BC, together with BC2. Upon BC describe (Prop. 46. 1.) the square CDEB, and produce ED to F, and through A draw (Prop. 31. 1.) AF parallel to CD or BE; then AE=AD +CE. But AE AB.BE = AB.BC, because BE BC. So also AD=AC. CD=AC.CB; and CE=BC2; therefore AB.BC=AC.CB+BC2. A C F D SCHOLIUM. B E In this proposition let AB be denoted by a, and the segments AC and CB, by b and c; then a=b+c: therefore, multiplying both members of this equality by c, we shall have ac=bc+c2. PROP. IV. THEOR. If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts. Let the straight line AB be divided into any two parts in C; the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB, that is, AB2=AC2+CB2+2AC.CB. B C G H K Upon AB describe (Prop. 46. 1.) the square ADEB, and join BD, and through C draw (Prop. 31. 1.) CGF parallel to AD or BE, and through G draw HK parallel to AB or DE. And because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal (29. 1.) to the interior and opposite angle ADB; but ADB is equal (5. 1.) to the angle ABD, because BA is equal to AD, being sides of a square; wherefore the angle CGB is equal to the angle GBC; and therefore the side BC is equal (6. 1.) to the side CG; but CB is equal (34. 1.) also to GK and CG to BK; wherefore the figure CGKB is equilateral. It is likewise rectangular; for the angle CBK being a right angle, the other. angles of the parallelogram CGKB are also right angles Wherefore CGKB is a square, and it is upon the side CB. D F E (Cor. 46. .) For the same |