BOOK IV.

PROPORTIONS, AREAS, AND SIMILARITY OF FIGURES.

DEFINITIONS.

209. The AREA of a figure is its quantity of surface, and is expressed by the number of times which the surface contains some other area assumed as a unit of measure.

Figures have equal areas, when they contain the same unit of measure an equal number of times.

210. SIMILAR FIGURES are such as have the angles of the one equal to those of the other, each to each, and the sides containing the equal angles proportional.

211. EQUIVALENT FIGURES are such as have equal areas. Figures may be equivalent which are not similar. Thus a circle may be equivalent to a square, and a triangle to a rectangle.

212. EQUAL FIGURES are such as, when applied the one to the other, coincide throughout (Art. 34, Ax. 14). Thus circles having equal radii are equal; and triangles having the three sides of the one equal to the three sides of the other, each to each, are also equal.

Equal figures are always similar; but similar figures may be very unequal.

213. In different circles, SIMILAR ARCS, SEGMENTS, or SECTORS are such as correspond to equal angles at the centres of the circles.

E

D

C

215. The ALTITUDE OF A PARALLELOGRAM is the perpendicular which measures the distance between its opposite sides taken as bases; as the perpendicular EF measuring the distance between the opposite sides, A B, D C, of the parallelogram ABCD.

A

216. The ALTITUDE OF A TRAPEZOID is the perpendicular distance between its parallel sides; as the distance measured by the perpendicular EF between the parallel sides, A B, D C, A of the trapezoid A B C D.

PROPOSITION I.

F

B

THEOREM.

lelogram be placed on that of the other, so that A B shall be the common base. Now, since the two parallelograms are of the same altitude, their upper bases, D C, FE, will be in the same straight line, DCEF, parallel to A B. From the nature of parallelograms DC is equal to A B, and FE is equal to A B (Prop. XXXI. Bk. I.); therefore DC is equal to FE (Art. 34, Ax. 1); hence, if DC and FE be taken away from the same line, D E, the remainders CE and DF will be equal (Art. 34, Ax. 3). But AD is equal to BC and AF to BE (Prop. XXXI. Bk. I.); therefore the triangles DAF, CBE, are mutually equilateral, and consequently equal (Prop. XVIII. Bk. I.).

If from the quadrilateral A B E D, we take away the triangle A D F, there will remain the parallelogram ABEF; and if from the same quadrilateral A B E D, we take away the triangle CBE, there will remain the parallelogram ABCD. Hence the parallelograms A B C D, ABEF, which have equal bases and equal altitude, are equivalent.

218. Cor. Any parallelogram is equivalent to a rectangle having the same base and altitude.

PROPOSITION II.-THEOREM.

219. If a triangle and a parallelogram have the same base and altitude, the triangle is equivalent to half the parallelogram.

Let A BE be a triangle, and D A BCD a parallelogram having the same base, A B, and the same altitude; then will the triangle be equivalent to half the parallelogram.

A

C F

E

B

Draw AF, FE so as to form the parallelogram ABEF. Then the parallelograms A B CD, ABEF, having the same base and altitude, are equivalent (Prop. I.). But

the triangle ABE is half the parallelogram ABEF (Prop. XXXI. Cor. 1, Bk. I.); hence the triangle ABE is equivalent to half the parallelogram ABCD (Art. 34, Ax. 7).

220. Cor. 1. Any triangle is equivalent to half a rectangle having the same base and altitude, or to a rectangle either having the same base and half of the same altitude, or having the same altitude and half of the same base.

221. Cor. 2. All triangles which have equal bases and altitudes are equivalent.

PROPOSITION III.-THEOREM.

222. Two rectangles having equal altitudes are to each other as their bases.

First. Suppose that the bases A B, A E are commensurable, and are to each other, for example, as the numbers 7 and 4. If A B is divided into seven equal parts, A E will contain four of those parts. At each point of division draw lines perpendicular to the base; seven rectangles will thus be formed, all equal to each other, since they have equal bases and the same altitude (Prop. I.). The rectangle ABCD will contain seven partial rectangles, while A EFD will contain four; hence the rectangle A B C D is to A EFD as 7 is to 4, or as AB is to A E. The same reasoning may be applied, whatever be the numbers expressing the ratio of the bases; hence, whatever be that ratio, when its terms are commensurable, we shall have

ABCD: AEFD::AB: AE.

Second. Suppose that the bases AB, D AE are incommensurable; we shall still have

ABCD: AEFD::AB: AE.

For, if this proportion be not true, the A first three terms remaining the same, the must be either greater or less than A E. to be greater, and that we have

ABCD: AEFD::AB: A O.

FK C

EIO B

fourth term Suppose it

Conceive A B divided into equal parts, each of which is less than EO. There will be at least one point of division, I, between E and O. Through this point, I, draw the perpendicular IK; then the bases A B, AI will be commensurable, and we shall have

ABCD:AIKD::AB: A I.

But, by hypothesis, we have

ABCD: AEFD::AB: AO.

In these two proportions the antecedents are equal; hence the consequents are proportional (Prop. X. Cor. 2, Bk. II.), and we have

AIKD: AEFD::AI: A O.

But AO is greater than AI; therefore, if this proportion is correct, the rectangle A EFD must be greater than the rectangle A IKD (Art. 125); on the contrary, however, it is less (Art. 34, Ax. 8); therefore the proportion is impossible. Hence, ABCD cannot be to AEFD as A B is to a line greater than A E.

In the same manner, it may be shown that the fourth term of the proportion cannot be less than A E; therefore it must be equal to A E. Hence, any two rectangles ABCD, AEFD, having equal altitudes, are to each other as their bases A B, A E.