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Because e is to fas (g to h, that is, as) y to z; and fis to g as (k to 1, that is, as) z to a: therefore, ex æquali, e is to g as y to a. And by the hypothesis, a is to b, that is, s is to t, as e to g; wherefore s is to t, as y is to a; and, by inversion, t is to s, as a to y; and s is to x, as y to d; therefore, ex æquali, t is to x, as a to d. Also, because h is to k as (c to d, that is, as) t to v; and k is to l, as (e to f, that is, as) v to x; therefore, ex æquali, h is to l, as t to x. In like manner, it may be demonstrated that m is to p, as a to d: and it has been shewn that t is to x as a to d; therefore (v. 11) h is to l, as m to p. Q. E. D.

The propositions G and K are usually, for the sake of brevity, expressed in the same terms with propositions F and H : and therefore it was proper to shew the true meaning of them when they are so expressed; especially since they are very frequently made use of by geometers.

EXERCISES ON BOOK V.

THEOREMS.

1. If three magnitudes have the same ratio to each other, the sum of the first and third is greater than twice the second.

2. The differences of magnitudes that are continual proportionals are proportionals also.

3. If there be four proportional magnitudes of which the first is a multiple of the second, then the third is the same multiple of the fourth. (Prove also that if the second be a multiple of the first, the fourth is the same multiple of the third.)

4. If the ratio between the first and second of four magnitudes to the second be greater than the ratio between the third and fourth to the fourth, then the first has a greater ratio to the second than the third has to the fourth.

5. If there be four proportional magnitudes, the ratio between the third and fourth is equal to, greater, or less, according as the first and second are equal to, greater, or less than each other.

BOOK VI.

DEFINITIONS.

I. SIMILAR rectilineal figures are those which have their several angles equal, each to each, and the sides about the equal angles proportionals.

Definition I.

II. Reciprocal figures, viz. triangles and parallelograms, are such as have their sides about two of their angles proportionals in such manner that a side of the first figure is to a side of the other as the remaining side of this other is to the remaining side of the first.

III. A straight line is said to be cut in extreme and mean ratio when the whole is to the greater segment as the greater segment is to the less. IV. The altitude of any figure is the straight line drawn from its vertex perpendicular to the base.

Definition IV.

PROPOSITION I.-THEOREM.

Triangles and parallelograms of the same altitude are one to another as

their bases.

LET the triangles a bc, a cd, and the parallelograms ec, cf, have the same altitude viz. the perpendicular drawn from the point a to bd: then, as the base bc is to the base c d, so is the triangle abc to the triangle a cd, and the parallelogram e c to the parallelogram cf.

Produce bd both ways to the points h, 1, and take any number of straight lines bg, gh, each equal to the base bc; and dk, kl, any number of them, each equal to the base cd; and join ag, ah, ak, al: then, because cb, bg, gh are all equal, the triangles ahg, a gb, abc, are all equal (i. 38): therefore, whatever multiple the base h c is of the

base bc, the same multiple is the triangle a hc of the triangle a bc. For the same reason, whatever multiple the base lc is of the base c d, the same multiple is the triangle alc of the triangle a dc and if the base hc be equal to the base cl, the triangle a hc is also equal to the triangle alc (i. 38): and if the base h c be greater than the base cl, likewise the triangle a hc is greater than the triangle alc; and if less, less therefore, since there are four magnitudes, viz. the two bases bc, cd, and the two triangles abc, acd; and of the base bc, and the triangle abc, the first and third, any equimultiples whatever have been taken, viz. the base hc

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and triangle a h c; and of the base c d and triangle a c d, the second and fourth, have been taken any equimultiples whatever, viz. the base cl and triangle alc; and that it has been shewn, that if the base hc be greater than the base cl, the triangle a hc is greater than the triangle alc; and if equal, equal; and if less, less therefore (v. def. 5), as the base bc is to the base cd, so is the triangle abc to the triangle a cd

And because the parallelogram ce is double of the triangle abc (i. 41), and the parallelogram cf double of the triangle a cd, and that magnitudes have the same ratio which their equimultiples have (v. 15); as the triangle abc is to the triangle a cd, so is the parallelogram e c to the parallelogram cf; and because it has been shewn, that as the base bc is to the base cd, so is the triangle a bc to the triangle acd; and as the triangle abc is to the triangle a cd, so is the parallelogram ec to the parallelogram cf; therefore, as the base bc is to the base cd, so is (v. 11) the parallelogram ec to the parallelogram cf. Wherefore, triangles, &c. Q. E. D.

COR. From this it is plain, that triangles and parallelograms that have equal altitudes are one to another as their bases.

Let the figures be placed so as to have their bases in the same straight line ; and having drawn perpendiculars from the vertices of the triangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are (i. 33), because the perpendiculars are both equal and parallel to one another. Then, if the same construction be made as in the proposition, the demonstration will be the same.

PROPOSITION II.—THEOREM.

If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or these produced, proportionally; and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle.

LET de be drawn parallel to bc, one of the sides of the triangle a bc: bd is to da, as ce to ea

Join be, cd; then the triangle bde is equal to the triangle cde (i. 37), because they are on the same base de, and between the same parallels de, bc. But a de is another triangle, and equal magnitudes have to the same the same ratio (v. 7); therefore, as the triangle bde to the triangle a de, so is the triangle cde to the triangle a de, but as the triangle bde to the triangle a de, so is (vi. 1) bd to da, because having the same altitude, viz. the perpendicular drawn from the point e to a b,

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they are to one another as their bases; and for the same reason, as the triangle cde to the triangle a de, so is ce to ea

da, so is ce to e a (v. 11).

Therefore, as bd to

Next, let the sides a b, a c, of the triangle a bc, or these produced, be cut proportionally in the points d, e, that is, so that bd be to da as ce to ea, and join de; de is parallel to bc

The same construction being made, because as bd to da, so is ce to ea; and as bd to da, so is the triangle bde to the triangle a de (vi. 1); and as ce to ea, so is the triangle cde to the triangle ade; therefore the triangle bde is to the triangle a de, as the triangle c d e to the triangle a de; that is, the triangles bde, cde have the same ratio to the triangle a de; and therefore (v. 9) the triangle bde is equal to the triangle cde; and they are on the same base de but equal triangles on the same base are between the same parallels (i. 39); therefore de is parallel to bc. Wherefore, if a straight line, &c. Q. É. D.

PROPOSITION III.-THEOREM.

If the angle of a triangle be divided into two equal angles by a straight line which also cuts the base, the segments of the base shall have the same ratio which the other sides of the triangle have to one another; and if the segments of the base have the same ratio which the other sides of the triangle

have to one another, the straight line drawn from the vertex to the point of section divides the vertical angle into two equal angles.

LET the angle bac of any triangle a b c be divided into two equal angles by the straight line ad: bd is to dc as ba to a c

e

a

Through the point c draw ce parallel (i. 31) to da, and let ba produced meet ce in e; because the straight line ac meets the parallels a d, ec, the angle ace is equal to the alternate angle cad (i. 29): but cad, by the hypothesis, is equal to the angle bad; wherefore bad is equal to the angle a ce. Again, because the straight line bae meets the parallels a d, ec, the outward angle bad is equal to the inward and opposite angle a ec: but the angle a ce has been proved equal to the angle bad; therefore also ace is equal to the angle a ec, and consequently the b side a e is equal to the side a c (i. 6):

d

c

and because ad is drawn parallel to one of the sides of the triangle bce, viz. to ec, bd is to dc, as ba to ae (vi. 2), but ae is equal to ac; therefore as bd to dc, so is ba to ac (v. 7).

Let now bd be to dc, as ba to a c, and join a d; the angle bac is divided into two equal angles by the straight line a d.

The same construction being made; because, as bd to d c, so is ba to ac; and as bd to cd so is ba to ae (vi. 2), because a d is parallel to ec; therefore ba is to a c, as ba to ae (v. 11); consequently ac is equal to a e (v. 9), and the angle aec is therefore equal to the angle ace (i. 5): but the angle a ec is equal to the outward and opposite angle bad; and the angle ace is equal to the alternate angle cad (i. 29): wherefore also the angle bad is equal to the angle cad; therefore the angle bac is cut into two equal angles by the straight line ad. Therefore, if the angle, &c. Q. E. D.

PROPOSITION A.-THEOREM.

If the outward angle of a triangle made by producing one of its sides be divided into two equal angles by a straight line which also cuts the base produced, the segments between the dividing line and the extremities of the base have the same ratio which the other sides of the triangle have to one another; and if the segments of the base produced have the same ratio which the other sides of the triangle have, the straight line drawn from the vertex to the point of section divides the outward angle of the triangle into two equal angles.

LET the outward angle cae of any triangle abc be divided into two qual angles by the straight line ad which meets the base produced in bd is to dc, as ba to a c.

Through c draw cf parallel to a d (i. 31); and because the straight line ac meets the parallels a d, cf, the angle a cf is equal to the alternate

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