Proposition 14.

Theorem. If four straight lines be proportional, the rectangle of the extremes is equal to the rectangle of the means; and if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportional.

A

Let A, B, C, D be four straight lines, so that A : B :: C : D; then the rectangle contained by A and D is equal to the rectangle contained by B and C.

Make EF equal to A, and produce it, making FG equal to B; through F draw HK at right angles to EG, making FH equal to C, and FK E equal to D, and complete the parallelograms EK, KG, GH.

Now (V. 1),
also (V. 1),
But (Hyp.)
Hence (IV. 15),

... (IV. 14),

B

C

D

K

H

F

G

EK : KG :: EF : FG;
HG:KG :: HF : FK.

EF : FG :: HF : FK.

EK : KG :: HG : KG;
EK=HG.

But EK is the rectangle contained by EF, FK, or by A and D, and HG is the rectangle contained by Band C. Therefore, the rectangle contained by A and D is equal to the rectangle contained by B and C.

Conversely, let EK be equal to HG. then A : B :: C :D. For, because EK = HG (IV. 14),

But (V. 1) and

... (IV. 15),

or

EK : KG :: HG : KG.
EK: KG :: EF : FG,
HG:KG :: HF : FK;
EF : FG :: HF : FK,
A: B:: C : D.

Proposition 15.

Theorem. Parallelograms which are mutually equiangular

are proportional to the products of their adjacent sides.*

Let AB, AC be two parallelograms, which have their angles respectively equal; then

F

D

A

Corollary. Two triangles which have one angle of each equal, are proportional to the products of the sides about the equal angles.

For, join FH, DG; then AFH and ADG, being halves of AC and AB, we have

AFH : ADG :: AF×AH : AD×AG.

Proposition 16.

Theorem. The area of a rectangle is equal to the product of its adjacent sides.

Let AC be a rectangle; its area is equal to the product of AB and BС.

* By "the product of their adjacent sides," is meant the product of the number of units of the same kind in those sides.

Construct another rectangle DF;

then (V. 15), AC: DF :: AB×BC: DE×EF.

Now suppose DE and EF to be each equal to the linear unit of measure; then DF will be the square, which is the unit of measure of surfaces. Hence the above proportion becomes

or

A

AC:1::AB×BC : 1,
AC=AB×BC.

Scholium. If AB and BC be commensurable, then divide them into equal portions, each of which is a linear unit, and draw lines parallel to the sides; then it is evident that the number of square units in the surface is the product of the number of linear units in the two sides.

A

C

F

BD

E

C

B

Corollary 1.- The area of any parallelogram is equal to the product of its base and altitude.

For it is equal to a rectangle of the same base and altitude (Ι. 33).

Corollary 2.- The area of a triangle is equal to one half the product of its base and altitude.

For a triangle is one half a rectangle of the same base and altitude (I. 35, Cor.).

Proposition 17.

Theorem. The area of a trapezoid is equal to the sum of the parallel sides, multiplied by one half the perpendicular distance between them.

Let AC be a trapezoid; its area is equal to the sum of AB and DC, multiplied by one half of AE.

A

B

Join DB; then the area of the triangle ADB is equal (V. 16, Cor. 2) to one half the product of AB and AE; also the triangle DBC is equal to one half the product of DC and AE. Hence the area of the whole trapezoid is equal to the sum of AB and DC multiplied by one half AE.

Scholium 1. The area of any irregular polygon may be found by dividing it into triangles, and measuring their bases and altitudes; or by measuring a long diagonal and the perpendicular distances from the other angles, and finding the area of the triangles and trapezoids formed.

DE

C

Scholium 2.-Having now shown that the rectangle contained by two sides is equal to the product of the linear units of the same kind in those sides, we can use the product of two lines instead of the rectangle contained by those lines, and vice versa. Hence the (IV. 12) will now be, If four straight lines be proportional, and also four others, the rectangles of the corresponding terms are proportional. The truth of the (V.14) might also have been established in a similar manner from the (IV. 1). The other proof is given to show that the algebraic and the geometrical methods lead to the same result; one by the use of abstract numbers or their equivalents, the other by considerations of magnitudes without regard to the unit which measures them. It may also be observed that AC2 hereafter may properly mean either the square described on AC, or the second power of the number of units in AC. It has heretofore meant only the former. Hence from (IV. 12, Cor. 2) we obtain, If four straight lines be proportional, the squares described on them will be proportional.

Proposition 18.

Problem.- Upon a given straight line to describe a polygon similar to a given polygon.

Let AB be a given straight line, and CDEF a given polygon; it is required to describe on AB a polygon similar to CDEF.

maining angle CFD (I. 32, Cor. 4); wherefore the triangle FCD is equiangular to the triangle GAB. Again, at the points G, B in the straight line GB make (I. 5) the angles HGB, HBG equal to the angles EFD, EDF, each to each; then will the angle H be equal to the angle E, and the triangle GBH be equiangular to the triangle FDE. And because the angle AGB is equal to the angle CFD, and the angle BGH to the angle EFD, therefore the whole angle AGH is equal to the whole angle CFE. For the same reason the angle ABH is equal to the angle CDE. Therefore, the polygon ABHG is equiangular to the polygon CDEF.

The sides about the equal angles will be proportional; for because the triangles GAB, FCD are similar,

(V. 4),

also

BA: AG:: DC : CF;

AG: GB:: CF : FD;

also in the triangles GBH, FDE,

GB : GH :: DF : FE;

... (IV. 13), AG:GH :: CF : FE.