The Elements of Plane and Solid Geometry

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.

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 A C2 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.

Join DF; and at the points A and B in the straight line AB make (I. 13) the angle BAG equal to the angle C, and the angle ABG equal to the angle CDF; therefore the remaining angle AGB is equal to the re

maining angle CFD (I. 30, 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 A GB 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,

In the same manner it may be proved

also (V. 4),

BH:: CD: DE;

GH: HB :: FE : ED.

Wherefore the polygons are equiangular, and have the sides about the equal angles proportional; they are therefore similar.

In the same manner if the polygon contained five or more sides we could apply other triangles to the lines of the polygon ABHG, and prove the resulting polygons similar.

Proposition 19.

Theorem. Similar triangles are to one another as

of their homologous sides.

Let ABC, DEF be similar triangles, having the angle B equal to the angle E, and let AB BC: DE : EF, SO that BC and EF are homologous sides; then ABC: DEF :: BC2 : EF2.

Because ABC, DEF have the angles at B and E equal

to each other, we have (V. 15, Cor.)

C E

ABC: DEF :: AB×BC: DE× EF;

also (Hyp.), AB: BC :: DE : EF.

Multiplying this, term by term, by the identical proportion

we have (V. 17, Sch. 2) AB BC: BC: DE EF: EF2; alternately,

AB BC: DE× EF :: BC2: EF2.

But we have proved above,

AB BC: DE × EF :: ABC: DEF.

Hence (IV. 15), ABC: DEF :: BC2 : EF2.

Proposition 20.

Theorem. Similar polygons may be divided into the same number of similar triangles, having the same ratio to one another that the polygons have; and the polygons are to each other as the squares of their homologous sides.

Let ABCDE, FGHKL be similar polygons, and let AB be homologous to FG; they may be divided into the same number of similar triangles, whereof each has to each, the same ratio that the polygons have;

and

ABCDEFGHKL :: AB2 : FG2.

Join BE, EC, GL, LH; and because ABCDE is similar to FGHKL, the angle BAE is equal to the angle GFL,

and (V., Def. 1) BA: AE :: GF: FL.

Wherefore (V. 6), the tri

that is, the sides about the equal angles EBC, LGH are proportional; therefore the triangle EBC is similar (V. 6) to the triangle LGH. For the same reason the triangles ECD, LHK are similar. Therefore the polygons are divided into the same number of similar triangles.

These triangles have to each other the same ratio the poly

gons have.

Because

(V. 19),

also (V. 19),

.·. (IV. 15), Similarly, ... also,