Two sides of one figure are said to be reciprocally proportional to two sides of another, when one of the sides of the first is to one of the sides of the other, as the remaining side of the 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.

1

Book VI.

a 38. 1.

PROP. I. THEOR.

TRIANGLES of the same altitude are to one another as their bases; and parallelograms of the same altitude are to one another as their bases.

Let the triangles ABC, ACD and the parallelograms EC, CF, have the same altitude; as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and so is the parallelogram EC to the parallelogram CF.

Produce BD both ways to the points H, L; take any number of straight lines BG, GH, each equal to the base BC, and any number DK, KL, each equal to the base CD; and join AG, AH, AK, AL. Because CB, BG, GH are all equal, the triangles AHG, AGB, ABC are all equal; therefore whatever multiple the base HC is of the base BC, the same multiple is the triangle AHC of the triangle ABC. For the same reason, whatever multiple the base LC is of the base CD, the same multiple is the triangle ALC of the triangle ADC. But if the base HC be equal to the base CL, the triangle

E A

F

AHC is equal to the

triangle ALC2; and if

the base HC be greater H G B C D
than the base CL, the

triangle AHC is greater than the triangle ALC; and if less, less. Now there are four magnitudes, 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 and the triangle AHC; and of the base CD and the triangle ACD, the second and fourth, any equimultiples whatever have been taken, viz. the base CL and the triangle ALC; and it has been shown that if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ALC; and if equal, equal;

and if less, less. Therefore as the base BC is to the base Book VI. CD, so is the triangle ABC to the triangle ACDb.

Because the parallelogram CE is double of the triangle b 5. Def. 5. ABC, and the parallelogram. CF double of the triangle c 41. 1. ACD, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CFd. But it has been d 15. 5. shown that as the base BC is to the base CD, so is the triangle ABC to the triangle ACD; therefore as the base BC is to the base CD, so is the parallelogram EC to the parallelogram CFe. Wherefore, triangles, &c. Q. E. D.

COR. Triangles, and also parallelograms, that have equal altitudes, are to one another as their bases.

e 11. 5.

Let the figures be placed so as to have their bases in the same straight line, and let perpendiculars be drawn from the vertices of the triangles to the bases; the straight line which joins the vertices is parallel to that in which their bases aref, f 33. 1. because the perpendiculars are both equal and parallels to one g 28. 1, another. Then, if the same construction be made as in the proposition, the demonstration will be the same.

PROP. II. THEOR.

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

Let DE be drawn parallel to BC, one of the sides of the triangle ABC; BD is to DA, as CE to EA.

Book VI.

a 37. 1.
b 7. 5.
c 1. 6.

d 11. 5.

e 9. 5.

f 39. 1.

Join BE, CD; then the triangle BDE is equal to the triangle CDE. But ADE is another triangle, therefore as the triangle BDE to the triangle ADE, so is the triangle CDE to the triangle ADE.. But, as the triangle BDE to the triangle ADE, so isc BD to DA; and as the triangle CDE to the triangle ADE, so is CE to EA. Therefore as BD to DA, so is CE to EAa.

Next, let the sides AB, AC of the triangle ABC, or these sides 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, as BD to DA, so is CE to EA, and as BD to DA, so is the triangle BDE to the triangle ADE; and as CE to EA, so is the triangle CDE to the triangle ADE; therefore the triangle BDE is to the triangle ADE, as the triangle CDE to the triangle ADE*; therefore the triangle BDE is equal to the triangle CDE. But they are on the same base DE, therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q. E. D.

* This proportion depends upon the principle, that ratios which are equal to equal ratios are equal to one another, which may be inferred from 11. 5.