Therefore, AB is divided at D externally in extreme and mean ratio.

298. Cor. If AB be denoted by m, and AC by x, proportion (3) of § 297 becomes

Multiplying by 4, and adding m2 to both members, 4x2 + 4 mx + m2 = 4 m2 + m2 = 5m2.

Extracting the square root of both members, 2 x + m = ± m√5.

Since a cannot be negative, we take the positive sign before the radical sign; then,

77. To inscribe in a given circle a triangle similar to a given triangle. (§ 261.)

(Circumscribe a ○ about the given ▲, and draw radii to the

vertices.)

78. To circumscribe about a given circle a triangle similar to a given triangle. (§ 262.)

BOOK IV.

AREAS OF POLYGONS

PROP. I. THEOREM.

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

Note. The words "rectangle,' 99 66

'parallelogram," "triangle," etc.,

in the propositions of Book IV., mean the amount of surface in the rectangle, parallelogram, triangle, etc.

Case I. When the bases are commensurable.

Given rectangles ABCD and EFGH, with equal altitudes. AB and EF, and commensurable bases AD and EH.

Proof. Let AK be a common measure of AD and EH, and let it be contained 5 times in AD, and 3 times in EH.

Drawings to AD and EH through the several points of division, rect. ABCD-will be divided into 5 parts, and rect. EFGH into 3 parts, all of which parts are equal.

(§ 114)

(2)

(?)

Given rectangles ABCD and EFGH, with equal altitudes AB and EF, and incommensurable bases AD and EH.

Proof. Divide AD into any number of equal parts, and apply one of these parts to EH as a unit of measure.

Since AD and EH are incommensurable, a certain number of the parts will extend from E to K, leaving a remainder KH < one of the equal parts.

Draw line KL 1 EH, meeting FG at L.

Then, since AD and EK are commensurable,

Now let the number of subdivisions of AD be indefinitely increased.

Then the unit of measure will be indefinitely diminished, and the remainder KH will approach the limit 0.

By the Theorem of Limits, these limits are equal.

(?)

300. Cor. Since either side of a rectangle may be taken

as the base, it follows that

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

PROP. II. THEOREM.

301. Any two rectangles are to each other as the products of their bases by their altitudes.

Given M and N rectangles, with altitudes a and a', and bases b and b', respectively.

Proof. Let R be a rect. with altitude a and base b'.

Then, since rectangles M and R have equal altitudes, they are to each other as their bases.

($ 299)

(1)

And since rectangles R and N have equal bases, they are to each other as their altitudes.

(?) (2)

302. The area of a surface is its ratio to another surface, called the unit of surface, adopted arbitrarily as the unit of measure (§ 179).

The usual unit of surface is a square whose side is some linear unit; for example, a square inch or a square foot.

303. Two surfaces are said to be equivalent (~), when their areas are equal.

304. The dimensions of a rectangle are its base and altitude.

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

Note. In all propositions relating to areas, the unit of surface (§ 302) is understood to be a square whose side is the linear unit.

a

M

N

b

Given a and b, the altitude and base, respectively, of rect. M; and N the unit of surface, i.e., a square whose side is the linear unit.

To Prove that, if N is the unit of surface,

area Ma × b.

Proof. Since any two rectangles are to each other as the products of their bases by their altitudes (§ 301),

But since N is the unit of surface, the ratio of M to N is the area of M.

(§ 302)

.. area Max b.

306. Sch. I. The statement of Prop. III. is an abbreviation of the following:

If the unit of surface is a square whose side is the linear unit, the number which expresses the area of a rectangle is equal to the product of the numbers which express the lengths of its sides.

An interpretation of this form is always understood in every proposition relating to areas.