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gles; therefore the remaining angles AEF, FGC are also equal (32. 1.): But AEF is a right angle, therefore FGC is a right angle, and AG is perpendicular to BC. Q. E. D.

Cor. The triangle ADE is similar to the triangle ABC. For the two triangles BAD, CAE having the angles at D and E right angles, and the angle at A common, are equiangular, and therefore BA:AD :: CA: AE, and alternately BA : CA:: AD: AE; therefore the two triangles BAC,DAE, have the angle at A common, and the sides about that angle proportionals, therefore they are equiangular (6. 6.) and similar. Hence the rectangles BA.AE, CA.AD are equal.

PROP. K. THEOR. If from any angle of a triangle a perpendicular be drawn to the opposite side or base ; the rectangle contained by the sum and difference of the other two sides, is equal to the rectangle contained by the sum and difference of the segments, into which the base is divided by the perpendicular. Let ABC be a triangle, AD a perpendicular drawn from the angle A on the base BC, so that BD, DC are the segments of the base; (AC+AB) (AC-AB)=(CD+DB) (CD-DB).

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From A as a centre with the radius AC, the greater of the two sides, describe the circle CFG: produce AB to meet the circumference in E and F, and CB to meet it in G. Then because AF=AC, BF=AB+AC, the sum of the sides; and since AE=AC, BE=AC,

-AB=the difference of the sides. Also, because AD drawn from the centre cuts GC at right angles, it bisects it; therefore, when the perpendicular falls within the triangle, BG=DG-DB=DC-DB= the difference of the segments of the base, and BC=BD+DC=the sum of the segments. But when AD) falls without the triangle, BG= DG+DB=CD+DB=the sum of the segments of the base, and BC =CD-DB=the difference of the segments of the base. Now, in both cases, because B is the intersection of the two lines FE, GC, drawn in the circle, FB.BE=CB.BG; that is, as has been shown, (AC+AB) (AC-AB)=(CD+DB) (CD-DB). Therefore, &c.

ELEMENTS

OF

GEOMETRY.

SUPPLEMENT.

BOOK I.

OF THE QUADRATURE OF THE CIRCLE.

DEFINITIONS.

A

I.
CHORD of an arch of a circle is the straight line joining the ex-

tremities of the arch; or the straight line which subtends the. arch.

II. The perimeter of any figure is the length of the line or lines, by

which it is bounded.

III.
The area of any figure is the space contained within it.

AXIOM. The least line that can be drawn between two points, is a straight

line: and if two figures have the same straight line for their base, that which is contained within the other, if its bounding line or lines be not any where convex toward the base, has the least perimeter.

COR. 1. Mence the perimeter of any polygon inscribed in a circle, is less than the circumference of the circle.

Cor. 2. If from a point two straight lines be drawn touching a cirele, these two lines are together greater than the arch intercepted between them; and hence the perimeter of any polygon described about a circle is greater than the circumference of the circle.

PROP. I. THEOR.

D

If from the greater of two unequal magnitudes there be taken away its half, and from the remainder its half: and so on; There will at length remain a magnitude less than the least of the proposed magnitudes.

Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken away its half, and from the remainder its half, and so on; there shall at length remain a magnitude less than C. A

For C may be multiplied so as, at length, to become greater than AB. Let DE, therefore, be a multiple of C, which is greater than AB, and K let it contain the parts DF, FG, GE, each equal to C. From AB take BH equal to its half, and from the remainder AH, take HK equal to its half, and so on, until there be as many divisions in AB as there are in DE: And let the divisions in AB be AK, KH, HB. And because DE is greater than AB, and EG taken from DE is not greater than its half, bút BH taken from AB is equal to its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and GF is not greater than the half of GD, but HK is equal to the half of HA; therefore, the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK is less than C. Q. E. D.

B

C

E

PROP. II. THEOR. V

Equilateral polygons, of the same number of sides, inscribed in circles, are similar, and are to one another as the squares of the diameters of the circles.

Let ABCDEF and GHIKLM be two equilateral polygons of the same number of sides inscribed in the circles ABD, and GHK; ABCDEF and GHIKLM are similar, and are to one another as the squares of the diameters of the circles ABD, GHK.

Find N and O the centres of the circles; join AN aud BN, as also GO and Ho, and produce AN and GO till they meet the eircumfer• ences in D and K.

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Because the straight lines AB, BC, CD, DE, EF, FA, are all equal, the arches AB, BC, CD, DE, EF, FA are also equal (28. 3.). For the same reason, the arches GH, HI, IK, KL, LM,MG are all equal, and they are equal in number to the others; therefore, whatever part the arch AB is of the whole circumference, ABD, the same is the arch GH of the circumference GHK. But the angle ANB is the same part of four right angles, that the arch AB is of the circumference ABD (33. 6.); and the angle GOH is the same part of four right angles that the areh GH is of the circumference GHK (33.6.) therefore the angles ANB, GOH are each of them the same part of four right angles, and therefore they are equal to one another. The isosceles triangles ANB, GOH are therefore equiangular (6. 6.), and the angle ABN equal to the angle GHO; in the same manner, by joining NC, OI, it may be proved that the angles NBC, OHI are

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equal to one another, and to the angle ABN. Therefore the whole angle ABC is equal to the whole GHI; and the same may be proved of the angles BCD, HIK, and of the rest. Therefore, the polygons ABCDEF and GHIKLM are equiangular to one another; and since they are equilateral, the sides about the equal angles are proportion; als; the polygon ABCD is therefore similar to the polygon GHIKLM (def. 1. 6.)." And because similar polygons are as the squares of their homologous sides (20. 6.), the polygon ABCDEF is to the palygon GHIKLM as the square of AB to the square of GH; but because the triangles ANB, GOH are equiangular, the square of AB is to the square of GH as the square of A N to the square of GO (4. 6.), or as four times the square of AN to four times the square (15. 5.) of GO, that is, as the square of AD to the square of GK (2. Cor.8. 2.). Therefore also, the polygon ABCDEF is to the polygon GHIKLM

square of AD to the square of GK; and they have also been shown to be similar. Therefore, &c. Q. E. D.

Cor. Every equilateral polygon inscribed in a circle is also equiangular: For the isosceles triangles, which have their common vertex in the centre, are all equal and similar; therefore, the angles at their bases are all equal, and the angles of the polygon are therefore also equal.

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PROP. III. THEOR.

The side of any equilateral polygon inscribed in a circle being given, to find the side of a polygon of the same number of sides described about the circle.

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B

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Let ABCDEF be an equilateral polygon inscribed in the circle ABD; it is required to find the side of an equilateral polygon of the same number of sides described about the circle.

Find G the centre of the circle; join GA, GB, bisect the arch AB in H; and through H draw KHL touching the circle in H, and meeting GA and GB produced in K and L; KL is the side of the polygon required.

Produce G7 to N, so that GN may be equal to GL; join KN, and from G draw GM at right angles to KN, join also HG.

Because the arch AB is bisected in H, the aogle AGH is equal to the angle BGH (27. 3.); and because KL touches the circle in H, the angles LHG, KHG are right angles (16.3.); therefore, there are two angles of the triangle HGK, equal to two angles of the triangle HGL, each to each. But the side GH is K

to these triangles; therelore they are equal (26. 1.), and GL is equal to ĜK. Again, in the triangles KGL, KGN, hecanse GN is equal to GL; and GK common, and also the angle LGK equal to the angle KGN; therefore the base KL is equal to the base KN (4. 1.) But because the triangle KGN is isosceles, the angle GKN is equal

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