LB (by Prop. 6. B. 2.); add to both the square of AL, and the rectangle under CB and BO, together with the squares of AL and LO, or the square of AO (by Prop. 47. B. 1.), will be equal to the squares of AL and LB, and therefore to the square of AB (by Prop. 47. B.§1.), or to the squares of AF and FB (by Prop. 47. B. I.); therefore taking away the equal squares AO and AF, the rectangle under CB and BO will be equal to the square of BF (by Ax. 3.)

COR. 1.-Hence if from any point without a circle, two right lines be drawn cutting the circle, the rectangles under them and their external segments will be equal, for each is equal to the square of a touching line.

COR. 2.-If from the same point two right lines be drawn to a circle, which touch the circle, they will be equal; for their squares are equal, because either of them is equal to the same rectangle; that is, the rectangle under any line drawn from the same point, terminating on the concave of the circle, and its external segment.

COR. 3.-If from the angle A of a triangle BAC a perpendicular AL be let fall on the opposite side, the rectangle under the sum BE and difference BD of the sides AB and AC, will be equal to the rectangle under the sum and difference of the segments BL and LC, intercepted between the perpendicular and the extremities of the side on which it falls.

For from the centre A, and with the less side AC as an interval, describe a circle; and since the right line

E

BE and BC cut the circle, the rectangle under EB and DB will be equal to the rectangle under CB and BO (by Cor. 1), but AE is equal to AC, and therefore EB is equal to the sum of AC and AB; but AD is also equal to AC, and therefore BD is equal to the difference between AC and AB; and therefore the rectangle under EB and DB is the rectangle under the sum and difference of of AB and AC; but because AL is perpendicular to CO, OL and LC are equal (by Prop. 3. B. 3.); and therefore when the perpendicular falls within the triangle, BO is equal to the difference of BL and LC, but BC is their sum; and when the perpendicular falls outside the triangle (as in the other figure), BO is equal to the sum of them, and BC the difference, and therefore the rectangle under BC and BO is the rectangle under the sum and difference of them BC and BO; and is equal to the rectangle under the sum and difference of the sides AB and AC.

If from a point (B) without a circle two right lines be drawn to it, either of them (BC) cutting it and the other (BF) meeting it, and if the rectangle under the cutting line and its external segment, be equal to the square of the line which meets the circle, the meeting line (BF) touches the circle, or is a tangent to it.

For draw from B the right line BQ touching the circle, and also draw EF and EQ. Therefore since the right line BC cuts the circle, and BQ touches it, the square of BQ will be equal to the rectangle under BC and BO

E

F

(by Prop. 36. B. 3.); but the square of BF is equal to the rectangle under BC and BO (by Hypoth.), and therefore the squares of BQ and BF are equal, and therefore the right lines themselves are equal (by Cor.

2. Prop. 46. B. 1.); now in the triangles EFB and EQB, the sides EF and FB, are equal to the sides EQ and QB, but the side EB is common, and therefore the angle EFB is equal to EQB (by Prop. 8. B. 1.); but the angle EQB is a right angle (by Prop. 18. B. 3.), and therefore the angle EFB is right, and therefore the right line BF touches the circle (by Prop. 10. B. 3.), and is therefore a tangent to it.

QUESTIONS

REFERRING TO THE THIRD BOOK.

How do you complete a circle, having only a segment of it given?

Prove the 22nd Proposition, by assuming numbers for each angle?

Upon what previous Propositions does the proof of the 20th Proposition depend?

By what Proposition can you erect a perpendicular at the extremity of a given line?

MULTIPLICATION OF ALGEBRA.

RULE.

First. When the quantities (or letters) are similar; multiply the numerical co-efficients, as in Arithmetic; then add together the indices of the similar quantities, placing their sum over the common letter or letters; and if the multiplier and multiplicand have like signs, place plus before the product, but if unlike, place minus before it.

Secondly. When the quantities are not similar; multiply the numerical co-efficients, as before, and write down the unlike quantities after their product, prefixing, as above, the required sign to the result.

In Ex. 1, we have a to be multiplied by + a, that is, the first power of a to be multiplied by itself, which is equal to a X a, or aa, or a2, joined by adding

H

the indices, according to the Rule; it must be remembered that a is the first power of a, therefore it has one for its index, thus a. Again, the multiplier and multiplicand having like signs, the product becomes plus. In Ex. 2, the product is similarly found, and the signs being similar to each other, produce plus, as in the last example. But in Ex. 3, the signs are not similar, therefore they make minus.

Exs. 4 and 5 are both the same, therefore the index of a being, the sum of the indices becomes 1, that is a1. Wherefore multiplication of similar quantities is performed by adding their indices, and prefixing the sum to the right hand of the common letter or letters.

The product of each of the Examples 6, 7, and 8, is found by the second part of the general Rule; and the product of the 9th is found by both first and second part; namely, by first adding the indices of the similar quantities, and then writing them down as above. Let it now be remembered that a letter standing without a co-efficient, 1 is understood, therefore a means la, and b, lb.