which interpreted geometrically gives If a segment (a+c) be divided into two parts (a, c), the rectangle on the segment and one of its parts (a) is equal to the sum of the square on that part (a2) and the rectangle on the two parts (ac). This is Euclid, Book II., Prop. 3. The truth of the geometric theorem is manifest from a proper figure. In 160°. Homogeneity.-Let a, b, c, d denote segments. the linear expressions a+b, a− b, etc., and in the rectangular expressions ab + cd, etc., the interpretations of the symbols + and are given in 28°, 29°, and 143°, and are readily intelligible. But in an expression such as ab+c we have no interpretation for the symbol + if the quantitative symbols denote line-segments. For ab denotes the area of a rectangle and c denotes a segment, and the adding of these is not intelligible in any sense in which we use the word "add.” Hence an expression such as ab+c is not interpretable geometrically This is expressed by saying that-An algebraic form has no geometric interpretation unless the form is homogeneous, i.e., unless each of its terms denotes a geometric element of the same kind. It will be observed that the terms "square,' " "dimensions," homogeneous," and some others have been introduced into Algebra from Geometry. 161°. Rectangles in Opposite Senses.-The algebraic term ab changes sign if one of its factors changes sign. And to be consistent we must hold that a rectangle changes sense whenever one of its adjacent sides changes sense. Thus the rectangles AB. CD and AB. DC are the same in extent of area, but have opposite senses. And for and AB. CD+AB. DC=0, the sum=AB(CD+DC), CD+DC=0. (156°) As the sense of a rectangle depends upon that of a linesegment there is no difficulty in determining when rectangles. are to be taken in different senses. The following will illustrate this part of the subject Let OA=OA' and OC=OC', and let B the figures be rectangles. B 3 1 C B + A + C' 2 1 □s OA. OC and OA'. OC have the A common altitude OC and bases equal in length but opposite in sense. Therefore OA.OC and OA'. OC are opposite in sense, and if we call □ОA. OC positive we must call □0A'. OC negative. Again, s OC. OA' and OC'. OA' have the common base OA' and altitudes equal in length but opposite in sense. Therefores OC. OA' and OC'. OA' are opposite in sense, and therefore s OA. OC and OA'. OC' are of the same sense. Similarly s OC. OA' and OC'. OA are of the same sense. These fours are equivalent to the algebraic forms : +a.+b=+ab, +a.-b= − ab, -a.+b=-ab, Ex. 1. ABCD is a normal quadrangle whose opposite sides meet in O, and OE, OF are altitudes of the As DOC and AOB respectively. The Qd. ABCD =ADOC-AAOB, =DC.OE□AB.OF. (141°) Now, let A move along AB to A' A (104°). Then O comes to O', F to F', DE BF' ·A' E to E', and O'E', O'F' become the altitudes of the As DO'C and A'O'B respectively. But O'E' and OE have the same sense, therefore DC.OE and DC. OE' have the same sense. Also, A'B is opposite in sense to AB, and O'F' is opposite in sense to OF. (156°, Ex. 3) AB. OF and A'B. O'F' have the same sense; ... Qd. A'BCD=ADO'C-AA'O'B; or, the area of a crossed quadrangle must be taken to be the difference between the two triangles which constitute it. 162°. Theorem.-A quadrangle is equal to one-half the parallelogram on its diagonals taken in both magnitude and relative direction. Qd. ABCD=1PQRS in both figures. This theorem illustrates the generality of geometric results when the principle of continuity is observed, and segments and rectangles are considered with regard to sense. Thus the principle of continuity shows that the crossed quadrangle is derived from the normal one (156°, Ex. 2) by changing the sense of one of the sides. This requires us to give a certain interpretation to the area of a crossed quadrangle (161°, Ex. 1), and thence the present example shows us that all quadrangles admit of a common expression for their areas. 163°. A rectangle is constructed upon two segments which are independent of one another in both length and sense. But a square is constructed upon a single segment, by using it for each side. In other words, a rectangle depends upon two segments while a square depends upon only one. Hence a square can have only one sign, and this is the one which we agree to call positive. Hence a square is always positive. 164°. The algebraic equation abcd tells us geometrically that the rectangle on the segments a and b is equal to the rectangle on the segments c and d. But the same relation is expressed algebraically by the form therefore, since a is a segment, the form is linear and cd denotes that segment which with a determines a rectangle equal to cd. 165°. The expression a2bc tells us geometrically that the square whose side is a is equal to the rectangle on the segments b and c. But this may be changed to the form a=√bc. Therefore since a is a segment, the side of the square, the form bc is linear. Hence the algebraic form of the square root of the product of two symbols of quantity is interpreted geometrically by the side of the square which is equal to the rectangle on the segments denoted by the quantitative symbols. 166°. The following theorems are but geometric interpretations of well-known algebraic identities. They may, however, be all proved most readily by superposition of areas, and thus the algebraic identity may be derived from the geometric theorem. 1. The square on the sum of two segments is equal to the sum of the squares on the segments and twice the rectangle on the segments. α b a a 2 ab (a+b)2 = a2+b2+2ab. 2. The rectangle on the sum and difference of two segments is equal to the difference of the squares on these segments. (a+b)(a - b) = a2 — b2. 3. The sum of the squares on the sum and on the difference of two segments is equal to twice the sum of the squares on the segments. (a+b)2 + ( a − b)2=2(a2+b2), a>b. 4. The difference of the squares on the sum and on the difference of two segments is equal to four times the rectangle on the segments. (a+b)2 - (a - b)2=4ab, a>b. EXERCISES. 1. To prove 4 of Art. 166°. Let AH-a and Hb B A a ab b b E F ab HB-b be the segments, so that AB is their sum. Through H draw HG || to BC, a side of the Make HG=α, square on AB. and complete the square FGLE, as in the figure, so that FG is Ka-b. Then AC is (a+b)2 and EG is (a - b); and their difference is the four rectangles AF, HK, CL, and DE; but these each have a and b as adjacent sides. (a + b)2 − ( a − b )2=4ab. 2. State and prove geometrically (a − b)2= a2+b2 —2ab. 3. State and prove geometrically (a+b)(a+c)= a2+ a(b+c)+bc. 4. State and prove geometrically by superposition of areas |