square of BC (by Prop. 47. B. 1), and therefore the squares of BC and AC are equal to twice the rectangle under AC and CF, together with the square of AB; therefore the square of AB is less than the squares of AC and CB, by twice the rectangle under AC and CF.

SCHOL. 1.-If the angle CAB be a right angle, the angles A and F coincide, and the rectangle under AC and CF will be the square of AC, but it appears in this case that the square of AB is less than the squares of AC and CB by twice the square of AC (by Prop. 47. B. 1.)

SCHOL. 2.-Hence, given the sides of any triangle (in numbers), we are able to find its area, for subtract the square of either of the sides AB, which is not the greatest, from the sum of the squares of the remaining sides AC and CB, and by either of these sides AC divide half the remainder, namely, the rectangle under AC and CF; subtract the square of the quotient CF, from the square of the remaining side CB, and the square root, BF, of the residue, multiplied into half the former divisor AC, gives the area of the triangle. COR. If from any

the squares of the sides

containing that angle (AB and AP) will be double of the squares of the bisecting line (AC), and of half the side opposite that angle (A.)

If the bisecting right line be perpendicular to the side, it is evident from Prop. 47. B. 1.

But if not, from the angle A draw to the opposite side the pependicular AF, and as one of the angles ACB and ACP is obtuse, let ACB be obtuse, and therefore the square of AB is equal to the squares of AC and of CB, together with twice the rectangle under BC and CF, (by Prop. 12. B. 2), but the angle ACP is acute, and therefore the square of AP, together with twice the rectangle under PC and CF, or because PC and BC are

equal (by Hypoth.), with twice the rectangle under BC and CF (by Prop. 13. B. 2), is equal to the squares of AC and of CP; therefore the squares of AB and AP with twice the rectangle under BC and CF, are equal to twice the square of AC, with the squares of BC and CP, and with twice the rectangle under BC and CF, take away from both twice the rectangle under BC and CF; and the squares of AB and AP are equal to twice the square of AC, with the squares of BC and CP, or with twice the square of BC, because BC and CP are equal.

PROPOSITION XIV. PROBLEM.

To construct a square equal to a given rectilineal figure (Z.)

Construct a rectangular parallelogram CI equal to Z (by Prop. 45. B. 1), and if the adjacent sides are equal, the proposition is done. If not, produce either side IA, and make the produced part AL

B

A

equal to the adjacent side AC (by Prop. 3. B. 1); bisect IL in O, and from the centre O, and with the interval OL, describe a semicircle LBI, and produce CA till it meets he periphery in B: the square described on AB, will be equal to the given rectilineal figure. For draw OB; and because IL is bisected in O, and cut unequally in A, the rectangle under IA and AL, together with the square of OA, shall be equal to the square of OL (by Prop. 5. B. 2), or of OB, which is equal to OL, and therefore to the square of OA and AB (by Prop. 47. B. 1); take away from both the square OA, and the rectangle under IA and AL, is equal to the square of AB, but the rectangle under IA and AL is equal to IC, for AL and AC are equal (by Constr.); therefore the square of AB, is equal to the rectangle IC, and therefore to the rectilineal figure Z.

SUBTRACTION OF ALGEBRA.

This Rule being the reverse of Addition, we change the signs of the quantities to be subtracted, or consider them as changed; that is, make negative signs affirmative, and affirmative signs negative; and then incorporate them as in Addition.

In Ex. 1, we have 2a to be subtracted from 5a, and therefore, proceeding according to the Rule, 2a becomes

2a; then, by subtracting the similar quantities having unlike signs, and prefixing the sign of the greater, as in Addition, we have 3a for the result.

Ex. 2 and 3 are combined in the same manner; but Ex. 4, being composed of unlike quantities, requires to be written down as in Addition, with the exception of changing the sign of the quantity to be subtracted.

The answers of Ex. 8, 9, 10, may be also written

Ex. 9, 10 a3c—d3⁄4.x

2. 10

In Ex. 9, it is evident that once the root of d.x taken from twice the root of d.x leaves once the root of d.x. And in Ex. 10, it is evident that the quantities being unlike, must be written as in addition, with the exception of changing the signs of the subtraction.

dex,

may

Parenthesis are used with much advantage, when any number of quantities are multiplied by the same co-efficient. Ex. gr., ax + cx be written thus, (a+cde); and any quantities having a common multiplier can be similarly written; which facilitates calculations considerably.

2.3 (ae) + x2 (b + n) + x2y (c-g)—m + p or ax3- ex3 + bx2 + nx2 + cx13y — gx2y — m + p

Whenever a negative sign stands before a parenthesis, it affects each sign within it; so that if we have, −(a + b − c)x, each of the signs contained within the parenthesis requires to be changed whenever we remove it, thus, —(a +b — c)x = ax bx + cx; and again, if we wish to place this in a parenthesis, it is also required that we should change the signs, thus, ax bx + cx becomes —(a + b — c)x.

(a — m).x13 + (p + n) x2 − (b + d)x + c + r