the arc an, their chords forming the inscribed polygon abed, &c. Circumscribe the circle with a similar polygon ABCD, &c., then this last will exceed the former by a magnitude less than the proposed magnitude M. From the centre O draw the lines Oa, OA, Oh, OH, and produce hO to d; then the polygon abcd, &c. is composed of as many triangles equal to Oah as the polygon has sides, and in like manner the polygon ABCD, &c. is composed of as many triangles equal to OAH as this polygon has sides, and as the polygons have each the same number of sides, the inscribed is the same multiple of the triangle Oah that the circumscribed is of the triangle OAH. Now the triangle Omh is half the triangle Oah, in like manner the triangle OhA is half the triangle OAH; hence the inscribed polygon is the same multiple of Omh that the circumscribed polygon is of OhA, and consequently,

Omh OhA ins. pol. circ. pol.,

:

whence (Prop. XIII. B. V.)

OhA: OhA- Omh:: circ. pol. : circ. pol.-ins. pol., that is,

OhA: Amh circ. pol.:: circ. pol.-ins. pol.

Now OhA is a right angle, and since AO bisects the angle A of the isosceles triangle aAh, it is perpendicular to ah; therefore the triangles OhA, Amh are similar, consequently, (Prop. XXI. B. VI.).

OhA: Amh:: Oh2: hm2 :: hd2: ha2, whence

hd ha2 circ. pol.:: circ. pol.-ins. pol.

:

Now a circumscribed square, that is to say, hd is greater than the polygon ABCD, &c., since the surfaces of circumscribed polygons diminish as their sides increase in number, so that in the last proportion the first antecedent is greater than the second, consequently, the first consequent is greater than the second, that is, the excess of the circumscribed polygon above the inscribed is less than ha2, and therefore less than N2 or than M.

Cor. As the circle is obviously greater than any inscribed polygon and less than any circumscribed one, it follows that a polygon may be inscribed or circumscribed, which will differ from the circle by less than any assignable magnitude.

PROPOSITION XI. THEOREM.

A circle is equivalent to the rectangle contained by lines equal to the radius and half the circumference. Let us represent the rectangle of the radius and semi-cir

cumference of the circle ABCD by P: we are to show that this rectangle is equal in surface to the circle.

If the rectangle P be not equivalent to the circle it must be either greater or less. Suppose it to be greater, and let us represent the excess by Q.

A

B

D

C

Then, by the corollary to last proposition, a polygon may be circumscribed about the circle, which shall differ therefrom by a magnitude less than Q, and must consequently be less than the rectangle P. But every circumscribed polygon is equivalent to the rectangle of the radius and half its perimeter, and the perimeter exceeds the circumference of the circle, consequently the rectangle of the radius of the circle and semi-perimeter of the polygon must be greater than P, the rectangle of the same radius and semi-circumference of the circle; but it was shown above to be less, which is absurd; hence the hypothesis that P is greater than the circle, is false.

But suppose the rectangle P is less than the circle, and let us represent the defect by the same letter Q.

Then, by the same corollary, a polygon may be inscribed in the circle, which shall differ from it by a magnitude less than Q, and must consequently be greater than the rectangle P. But every inscribed polygon is equivalent to the rectangle of its apothem and half its perimeter, and the apothem is less than the radius of the circle, and the perimeter less than the circumference; consequently the rectangle of the apothem and semi-perimeter of the polygon must be less than P, the rectangle of the radius and semi-circumference of the circle; but it was shown above to be greater, which is absurd; hence the second hypothesis also is false.

As therefore the circle can be neither greater nor less than the rectangle P, it must necessarily be equivalent to it.

Circles are to each other as the squares of their radii. Let the circles ABCD, abcd, be compared, we shall have the proportion

AO ao circ. ABCD : circ. abcd.

For if this proportion has not place let there be

AO2 ao :: circ. ABCD : P,

P being some magnitude either greater or less than the circle abcd. Suppose it to be less, and let us represent the defect by Q. Then (Prop. X. Cor.) a polygon may be inscribed in the circle abcd, which shall differ from it by a magnitude less than Q, and will therefore exceed the magnitude P. Let abcde, &c. be such a polygon, and describe a similar polygon ABCDE, &c. in the other circle. Then (Prop. V.)

AO: ao2:: pol. ABCDE, &c. : pol. abcde, &c. combining this with the proportion above we have

circ. ABCD: P:: pol. ABCDE, &c. : pol. abcde, &c. Now in this proportion the first antecedent is greater than the second, consequently the first consequent is greater than the second, that is, P is greater than the polygon abcde, &c., but it has been shown to be less, which is absurd. Therefore P cannot be less than the circle abcd.

But suppose that P is greater than the circle abcd. Then, still representing the difference by Q, a polygon may be circumscribed about the circle abcd, which shall differ from it by a magnitude less than Q, or be less than P. Suppose such a polygon to be described, and that a similar one is formed about the circle ABCD, then these polygons being to each other as the squares of the radii of their respective circles, it will evidently result by combining, as in the preceding case, this proposition with that advanced in the hypothesis, that the circle ABCD is to P, as the polygon about this circle to the polygon about the other; in which proportion the first antecedent is less than the second, and consequently the first consequent is less than the second, that is, P is less than the polygon circumscribed about the circle abcd; but it was shown above to be

greater, which is impossible. Hence, P can neither be less nor greater than the circle abcd, consequently it must be equal to it, and therefore,

AO2 ao2:: circ. ABCD : circ. abcd.

Cor. 1. Since every circle is equivalent to the rectangle of its radius and half its circumference, the above proportion may be expressed thus:

AO: ao: AO ABCD: ao abcd,

whence (Prop. I. Cor. B. VI.)

AO ao ABCD: abcd.

Consequently the circumferences of circles are to each other as their radii, and therefore their surfaces are as the squares of the circumferences.

Cor. 2. It follows also, that similar arcs are to each other as the radii of the circles to which they belong, for they subtend equal angles at the centres (Def. 2. B. VI.) and each angle is to four right angles as the arc which subtends it is to the whole circumference (Prop. XXIII. Cor. 1. B. VI.); consequently the one arc is to the whole circumference, of which it forms part, as the other arc to the circumference of which it is part: and as the circumferences are as the radii, we have alternately the one arc to the other as the radius of the former to that of the latter.

Cor. 3. Therefore also similar sectors are to each other as the squares of their radii, for each sector is to the circle as the arc to the circumference (Prop. XXIII. Cor. 2. B. VI); consequently the one sector is to its circle as the other sector to its circle: and as the circles are as the squares of the radii, we have alternately the one sector to the other as the square of the radius of the former to the square of that of the latter.

Cor. 4. It readily follows that similar segments are also as the squares of the radii, for they result from similar sectors, by taking away from each the triangle formed by the chord and radii, which triangles being similar, are also to each other as the squares of the radii; therefore the sectors and triangles being proportional, it follows (Prop. XI. B. V.) that the segments also are as the sectors or as the squares of the radii, or indeed as the squares of their chords.

Scholium.

From this proposition and its corollaries may easily be derived an extension of the property of the right angled triangle which forms proposition XXII. of Book VI., for it may now be

K

proved that if circles be described about the three sides taken as diameters, or if similar sectors or segments be formed on the sides, it will always result that the figure on the hypothenuse will be equivalent to both those on the sides. By turning to the proposition alluded to we shall find that the reasoning there employed applies equally to the demonstration of this property, and therefore it need not be here repeated. Another very remarkable property arises out of that just mentioned, and which must not remain unnoticed. Let semicircles be described on the hypothenuse BC, and on the sides AB, AC of the right angled triangle ABC; then since the semicircle BDAC is F equivalent to both the semicircles BFA, AGC, it follows that if the common segments BDA, AEC be taken away, there will remain the triangle ABC equivalent to the two circular spaces or lunes BFAD, AGCE.

B

G

E

D

PROPOSITION XIII. PROBLEM.

The surface of a regular inscribed polygon and that of a similar circumscribed polygon being given, to find the surfaces of regular inscribed and circumscribed polygons of double the number of sides.

Let ab be a side of the given inscribed polygon, then the tangents aA, bA will each be half the side of the similar circumscribed polygon: the chords a M,bM to the middle of the arc aMb will be sides of an inscribed polygon of double the number of sides; and lastly, the tangent BMC will be the side of a circumscribed

polygon similar to

this last. All this is evident from proposition IX.