PROPOSITION III. THEOREM.

Rectangles having equal altitudes, are proportional to their bases.

There may be two cases: the bases may be commensurable, or they may be incommensurable.

1o. Let ABCD and HEFK, be two rectangles whose altitudes AD and HK are equal, and whose bases AB and HE are commensurable: then will the areas of the rectangles be proportional to their bases.

Suppose that AB is to HE,

as 7 is to 4. Conceive AB to be divided into 7 equal parts, and HE into 4 equal parts, and at the points of division, let perpendiculars be drawn to AB and HE. Then will ABCD be divided into 7, and HEFK into 4 rectangles, all of which will be equal, because they have equal bases and equal altitudes (P. I.): hence, we have,

ABCD : HEFK :: 7 : 4.

But we have, by hypothesis,

AB : HE ::

7 : 4.

From these proportions, we have (B. II., P. IV.),

ABCD : HEFK :: AB : HE.

Had any other numbers than 7 and 4 been used, the same proportion would have been found; which was to be proved.

D

FK C

2o. Let the bases of the rectangles be incommensurable : then will the rectangles be proportional to their bases. For, place the rectangle HEFK upon the rectangle ABCD, so that it shall take the position AEFD. Then, if the rectangles are not proportional to their bases, let us suppose that

ABCD : AEFD :: АВ

A

E TO B

: A0;

Divide AB into

in which 40 is greater than AE. equal parts, each less than OE; at least one point of division, as I, will fall between E and 0; at this point, draw IK perpendicular to AB. Then, because AB and AI are commensurable, we shall have, from what has just been shown,

ABCD : AIKD :: AB AI.

:

The above proportions have their antecedents the same in each; hence (B. II., P. IV., C.),

AFFD : AIKD :: AO : AI.

The rectangle AEFD is less than AIKD; and if the above proportion were true, the line 40 would be less than AI; whereas, it is greater. The fourth term of the proportion, therefore, cannot be greater than AE. In like manner, it may be shown that it cannot be less than AE; consequently, it must be equal to AE: hence,

ABCD : AEFD :: AB AE;

which was to be proved.

Cor. If rectangles have equal bases, they are to each other as their altitudes.

Any two rectangles are to each other as the products of

their bases and altitudes.

H

D

Let ABCD and AEGF be two rectangles: then will ABCD be to AEGF, as AB × AD is to AE × AF For, place the rectangles so that the angles DAB and EAF shall be opposite or vertical; then, produce the sides CD and GE till they meet in H. The rectangles ABCD and ADHE have the same altitude

AD hence (P. III.),

B

IA

F

ABCD : ADHE :: AB : AE.

The rectangles ADHE and AEGF have the same altitude AE: hence,

ADHE : AEGF :: AD : AF

Multiplying these proportions, term by term (B. II., P. XII.), and omitting the common factor ADHE (B. II., P. VII.), we have,

ABCD : AEGF :: AB × AD :

AE × AF;

which was to be proved.

Scholium 1. If we suppose AE and AF, each to be equal to the linear unit, the rectangle AEGF will be the superficial unit, and we shall have,

ABCD : 1 :: AB × AD

1;

ABCD AB × AD:

hence, the area of a rectangle is equal to the product of its base and altitude; that is, the number of superficial units in the rectangle, is equal to the product of the number of linear units in its base by the number of linear units in its altitude.

Scholium 2. The product of two lines is sometimes called the rectangle of the lines, because the product is equal to the area of a rectangle constructed with the lines as sides.

PROPOSITION V. THEOREM.

The area of a parallelogram is equal to the product of its base and altitude.

Let ABCD be a parallelogram, AB its base, and BE its altitude: then will the area of ABCD be equal to AB x BE.

For, construct the rectangle. ABEF, having the same base and altitude : then will the rectangle be equal to the parallelogram (P. I.); but the area of the rectangle is equal to AB × BE:

F D

E C

hence, the area of the parallelogram is also equal to AB × BE; which was to be proved.

Cor. Parallelograms are to each other as the products of their bases and altitudes. If their altitudes are equal, they are to each other as their bases. If their bases are equal, they are to each other as their altitudes.

PROPOSITION VI. THEOREM.

The area of a triangle is equal to half the product of its

base and altitude.

Let ABC be a triangle, BC its base, and AD its altitude: then will the area of the triangle be equal to BC × AD.

E

For, from C, draw СЕ parallel to BA, and from A, draw AE parallel to CB. The area of the parallelogram BCEA is BC x AD (P. V.); but the triangle ABC is half of the parallelogram BCEA: hence, its area is equal to BC × AD; which was to be proved.

B

Cor. 1. Triangles are to each other, as the products of their bases and altitudes (B. II., P. VII.). If their altitudes are equal, they are to each other as their bases. If their bases are equal, they are to each other as their altitudes.

.Cor. 2. The area of a triangle is equal to half the product of its perimeter and the radius of the inscribed circle. For, let DEF be a circle

inscribed in the triangle ABC. Draw OD, OE, and OF, to the points of contact, and OA, OB, and OC, to the verti

ces.

D

B

area of OAC will be equal to OF × AC; and the area