for if we add the rectangle ab to the rectangle ac, so that a side a in the one shall coincide with an equal side a in the other, the sum makes a rectangle whose base is b + c and whose altitude is a; that is, the rectangle a(b + c).

In the same way, by adding three rectangles of the same altitude, we get

If there be any two sects one of which is divided into any number of parts, the rectangle contained by the two sects is equivalent to the rectangles contained by the undivided sect and the several parts of the divided sect.

If a sect be divided into any two parts, the rectangles contained by the whole and each of the parts are together equivalent to the square on the whole sect.

293. If c = a, then

a(b + c) = a(b + a) = ab + aa = ab + a2.

If a sect be divided into any two parts, the rectangle contained by the whole and one of the parts is equivalent to the rectangle contained by the two parts, together with the square on the aforesaid part.

294. The rectangle of two equal sects is a square, and (a + b)2 is only a condensed way of writing (a + b)(a + b). But, by the distributive law,

(a + b) (a + b) = (a + b)a + (a + b)b.

By the commutative law,

By the associative law,

a2 + ab + (ab + b2) = a2 + (ab + ab) + b2 = a2 + zab + b2,

Therefore

2

:. (a + b)2 = a2 + 2ab + b2.

If a sect be divided into any two parts, the square on the whole sect is equivalent to the squares on the two parts, together with twice the rectangle contained by the two parts.

295. COROLLARY. The square on a sect is four times the square on half the sect.

a(a + b + b) + b2 = a2 + ab + ab + b2 = (a + b)2.

From this, if we bisect the sect AB at C, and divide it into two unequal parts at D, and for BD put a, and for DC put b, we get the theorem :

If a sect is divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equivalent to the square on half the sect.

297. If CD = b, and AC = a + b, then their sum AD = a +b+b, and their difference is a; therefore the rectangle contained by their sum and difference equals a(a + 26), which, by 296, is the difference between (a + b)2 and b2.

Therefore

The rectangle contained by the sum and difference of any two sects is equivalent to the difference between the squares on those

sects.

298. If we bisect the sect AB at C, and produce it to D, and for BD put a, and for BC put b, then the above equation, a(a + b + b) + b2 = (a + b)2, gives us the theorem:

If a sect be bisected, and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together with the square on half the sect bisected, is equivalent to the square on the line which is made up of the half and the part produced.

By the associative law,

a2 + zab + b2 + a2 = 2a2 + zab + b2.

By the distributive law,

2a2 + 2ab + b2 = 2a (a + b) + b2.

By the commutative law,

Therefore

za (a + b) + b2 = 2(a + b) a + b2,
za(a

b)a

:. (a + b)2 + a2 = 2 (a + b) a + b2.

If a sect be divided into any two parts, the squares on the whole line and on one of the parts are equivalent to twice the

rectangle contained by the whole and that part, together with the square on the other part.

300. By the commutative and distributive laws, and 294,

4(a + b) a + b2 = 4a2 + 4ab + b2 = (2a + b)2.

By the associative and commutative laws,

(2a + b)2 = (a + a + b)2 = ([a + b] + a)2,

:. 4(a + b)a + b2 = ([a + b] + a)2.

If a sect be divided into any two parts, four times the rectangle contained by the whole sect and one of the parts, together