then will AFGD be a parallelogram of the same altitude with the trapezoid ABCD, having its base AF equal to the sum of the parallel sides of the trapezoid. It now remains to show that the trapezoid ABCD is equivalent to half the parallelogram AFGD. Since parallelograms of equal bases and altitudes are equivalent, (B. II, Prop. 1, Cor. 2,) we have the parallelogram DK equivalent to HF; also since the figure KH is a parallelogram, the triangle KBC is equal to the triangle BHC, (B. I, Prop. XVII ;) therefore the line BC divides the parallelogram AG into two equal parts, and the trapezoid ABCD is equivalent to half the parallelogram AG.

PROPOSITION V.

THEOREM. The square of the sum of two lines is greater than the sum of their squares, by twice the rectangle of the said lines; or, the square of a whole line is equal to the squares of its two parts, together with twice the rectangle of those parts.

any

Let the line AB be the sum of two lines AC, CB; then will the square of AB be equal to the sum of the squares of AC and CB, together with twice the rectangle AC.CB. That is,

For, let ABDF be the square on the sum or whole line AB, and ACGH the square on the part AC: produce CG and HG to meet the other sides at K and L. From

the lines CK, HL, which are equal, being each equal to a side of the square AB or BD, (B. I, Prop. xxv11,) take the parts CG, HG, which are also equal, being the sides of the square AG, and there remains GK equal to GL, which are also equal to LD and KD, being the opposite sides of the parallelogram GLDK; hence, the figure KL is equilateral, and it has all its angles right; (B. I, Prop. XXVII, Cor. 1;) it is therefore a square on the GL, or the square of its equal line CB. Also the figures FG, GB are equal to two rectangles under AC and CB, because HG is equal to AC, and GK or GL equal to CB. But the whole square AD is made up of four figures, namely, the two squares AG, GD, and the two equal rectangles FG, GB; that is, the square of AB is equal to the sum of the squares of AC and CB, together with twice the rectangle AC.CB.

Cor. If AC is equal to CB, we shall have AB'=4AC2; that is, the square of a line is equal to four times the square of half the line.

(45.) The algebraic formula (a+b)2=a2+2ab+b2, is equivalent to the foregoing proposition.

Thus, let AC=a; CB=b; and consequently AB=a+b.

As an example in numbers, suppose a=6; b=4; and we find

(6+4)==10262+2.6×4+42=36+48+16=100.

(46.) This theorem may be generalized, so as to apply in the case of a line which is the sum of any number of parts.

Let the line AB be equal to the sum of the four lines a, b, C, d; then will the square upon AB be made up of a2, b2, c2, d2, 2 ab, 2 ac, 2 ad, 2 bc, 2 bd, 2 cd.

This may be extended to the case where AB is composed of any number of parts, as follows, to wit:

The square of the sum of any number of lines, is equal to the sum of their squares increased by twice the rectangle of every two.

(47.) Another method of denoting the square of the sum of any number of lines, is as follows:

The square of the sum of any number of lines is equal to the square of the first line, plus twice the rectangle of the first line into the second; plus the square of the second, plus twice the rectangle contained by the sum of the first two into the third; plus the square of the third, plus twice the rectangle of the sum of the first three into the fourth; plus the square of the fourth, and so on. This becomes obvious by simply inspecting the following diagram :

PROPOSITION VI.

THEOREM. The square of the difference of two lines is less than the sum of their squares, by twice the rectan gle of the said lines.

Let AC, BC be any two lines, and AB their difference; then will the square of AB be less than the sum of the squares of AC, BC, by twice the rectangle of AC and BC. Or,

AB2=AC2+BC2-2AC.BC.

D

A

B

L M

For, let ABDF be the square on the difference AB, and ACGH the square on the line AC. Produce FD to K; also produce DB and KC, and draw LM, making BM the square of the line BC. Now it is obvious that the square AD is less than the two squares AG, BM, by the two rectangles FG, DM; but HG is equal to the one line AC, and FH or GK is equal to the other line BC; consequently the rectangle FG, contained under FH and HG, is equal to the rectangle of AC and BC. Again, GK being equal to CM, by adding the common part KC, the whole KM will be equal to the whole GC, or equal to AC; and consequently the figure DM is equal to the rectangle of AC and BC. Hence, the two figures FG, DM, are each rectangles of the two lines AC, BC;

and consequently the square of AB is less than the sum of the squares of AC, BC, by twice the rectangle AC.BC.

(48.) The algebraic formula (a-b)2=a2−2 ab+b2, is equivalent to this proposition.

Thus, let AC=a; BC=b; and consequently AB=a-b.

As an example in numbers, let a=10; b=6; and we find (10-6)2=42=10o-2.10×6+62=100-120+36=16.

PROPOSITION VII.

THEOREM. The rectangle under the sum and difference of two lines, is equal to the difference of the squares of those lines.

For, let ABDF be the square of AB, and ACGH the square of AC. Produce DB till BK be equal to AC; draw KL parallel to AB or FD, and produce GC both ways to L and M. Then the difference of the two squares AD, AG, is evidently the two rectangles FG, MB. But the rectangles FG, BL are equal; for FM and BK are each equal to AC, and HF is equal to CB, being equal to the difference between AB and AC, or their equals AF and AH. Therefore, the two rectangles FG, MB are equal to the two BL, BM, or to the whole MK; and consequently MK is equal to the difference of the squares AD, AC. But MK is a rectangle contained by DK the sum of AB and AC, and MD the difference of AB and