In the same manner it may be shown that the triangle BAD to the triangle DAC are similar to each other.

Wherefore in a right-angled triangle, etc., QED.

Corallary from this it is manifest that the perpendicular drawn from the right-angle of a right-angled triangle to the base, is a mean proportional between the segments of the base, and also that each of the sides is a mean proportional between the base and the segment of the base adjacent to that side.

For in the triangles BAD, BDC, AB is to BD as BD is to BC; [VI. 4.] And in the triangles DAC, BAD, AC is to AD as to AD is to AB; [VI. 4.]

and in the triangles DAC, BDC, AC is to CD as CD is to CB.

PROPOSITION 6. THEOREM.

[VI. 4.]

If three straight lines be proportionals, the rectangle contained by the extremes is equal to the square on the mean; and if the rectangle contained by the extremes is equal to the square on the mean, the three straight lines are proportionals.

See Plate 7 Fig. 2.

Let the three straight lines ABC be proportionals, namely: let A be to B as B is to C; the rectangle contained by A and C shall be equal to the square on B. Take D equal to B. Then, because A is to B as B is to C, [Hypothesis.]

and that B is equal to D; therefore A is to B as D is to C. [V. 7.] But if four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means; [VI. 16.] therefore the rectangle contained by A and C is equal to the rectangle contained by B and D. See Plate 7 Fig. 3 and 4. But the rectangle by B and D is the square on B, because B is equal to D; [Construction.] therefore the rectangle contained by A and C is equal to the square on B. Next, let the rectangle contained by A and C be equal to the square on B; A shall be to B as B is to C. For, let the same construction be made, then, because the rectangle contained by A and Cis equal to the square on B,

[Hypothesis.]

and that the square on B is equal to the rectangle contained by B and D, because B is equal to D;

6

therefore the rectangle contained by A and C is equal to the rectangle contained by B and D; but if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportionals; therefore A is to B as D is to C.

But B is equal to D;

Therefore A is to B as B is to C.

Wherefore if three straight lines, etc., QED.

QUADRATURE OF THE CIRCLE.

CASE 1.

[VI. 16.]

[Construction.]

[V.7.]

In which the Inscribed and Circumscribed Polygons are Carried to 22 Sides.

It is required to find the quadrature of the circle, that is when the radius is known it is required to find a straight line (in terms of the given radius) which shall be equal in length to the circumference of the given circle.

[Postulate 5.]

For this purpose let the square root of two be taken for the given radius fh, fm, and with this radius describe the given circle ADBC. [Postulate 3.]

See Plate 6 Fig. 1 and 2. Then let the square ABCD be inscribed in the given circle, the side of the inscribed square will be two and the area four.

[Hypothesis.]

For if each of the sides of the inscribed square ABCD be bisected the whole square will be divided into four smaller squares, each of the sides of which as well as their respective areas will be equal to one, consequently they will be equal to one another. [Axiom 1.]

Then, by (PROPOSITION 1, COR.), the diagonals of each of the smaller squares will be equal to the square root of two. But the radius of the given circle is equal to the square root of two, consequently the diagonals of each of the smaller inscribed squares is equal to the radius of the given circle. [Axiom 1.]

Again, all the radii are equal and each is half the diameter; [Definition 3.] therefore the diameter of the given circle is twice the square root of two, or the square root of eight, and it is equal to the diagonal of the square whose side is equal to the side of the inscribed square. For the half of the diameter of the given circle is the 1/2, and the half of

the side of the inscribed square is 1, consequently, by (Theorem 1, Cor.), the side of any square is to its diagonal as 1: √2; [Axiom 7.] therefore the one half of the side of the inscribed square is the side of a square whose diagonal is the square root of two, which is equal to the radius of the circle or half the diameter.

49 5

Now, by (PROPOSITION 2), the sine of the given arc is the square root of the th of the square described upon the radius, and the cosine of the given arc is the square root of the remaining 48. For, by trigonometry, R2. cos.2, and R2 cos.2 sin.2 quently cos.2+ sin.2 R2. Substituting the numbers we have of 2.04

(1/2)22.00 and

sin.2

R2.

50

conse

and the

and the .04

2 or

sine of the given are; and 2.00-.041.96 and the 1.96 and the 1/1.96=1.4 13 or cosine of the given arc.

f

2

25

49

Again, if the side of the small inscribed square, which is equal to 1, be divided into five equal parts each part is equal to , which is equal to the sine of the given arc; and seven of the same parts is equal to 7, which is equal to the cosine of the given arc; consequently (3)2=2 and (3)2=48, then 25+18=50 which is equal to the square described upon the radius; and the 1/50 1/2 which is equal to the radius. Now, by Theorem 3, the sine is not quite though very near the 4th part of the entire circumference; and the cosine is not quite though very near the entire diameter; therefore X 4444the assumed circumference; and X 214 the assumed diameter; 7 × 5 then, dividing the assumed circumference by the assumed diameter, we 44 14 44 筋 22 have by cancellation

5 5 $ 14 7

3.142857 or 34=

the true ratio between the assumed circumference and the assumed diameter of the given circle.

Again, the cosine of the given arc is 7, and the sine of the given arc is §, then, by trigonometry, cos.: sine :: R: tang. Substituting the numbers we have cos. : sin. :: radius √2: : tan. ‡ √/2, for } ×

1

√2=5√2; then, by division and cancellation, √2- 5

Again, by trigonometry, R2 + tan.2 sec.2. Substituting the numbers we have rad. (1/2)2= 2 and tang. (1/2)2=, then 2+ 400

and the √ 100=10=1.4142857, or = 1.44 secant of the given arc. =1.4: cosine of the given arc. = .2 1.2=sine of the given arc.

1/2 = .2020305089 tangent of the given are.

Again, the side of the inscribed square being divided into five parts, by applying the sine to it as a common measure each part is expressed by ; then X 1 the size of the square which is to be the first unit of superficial measure for the circle.

23

5

For, by ARTICLE 6, the area of the circumscribed double triangle ƒPE, f PE is equal to the product of the radius and the tangent; thus, radius 1/2 tangent 4/24; and, by ARTICLE 9, the area of the entire circumscribed polygon is equal to the product of the circumscribed double triangle, and the number of sides contained in the given polygon; thus & × 22 44 64 the area required. 44 25 1100 X 7 1

Now,

44

1

25

1574; therefore the area of

7=64;

the entire circumscribed polygon is equal to 1574 25

1100 44

175

that is, there is 1574 blocks each of which is equal in area to, QED.

CASE 2.

In which the Inscribed and Circumscribed Polygons are Carried to 3114 Sides.

By Case 1, the sine is and the double of the cosine is 14, and, by ARTICLE 3, the sine is a mean proportional between the double of the cosine and the tangent No. 2. For, by (PROPOSITIONS 4 and 5), sin. (1)2=, and cos. 214, then, by division and cancellation, 7 × 1 14 1 $ 1

2

25:5་

251

Х

25 14 70

tangent No. 2. (By PROPOSITION 6),

the rectangle contained by the double of the cosine and the tangent No. 2 is equal to the rectangle contained by the square described up

on the sine of the given arc.

For put double cosine

and put sine

and put tangent No. 2

A

B

Ꮯ

Then, by proportion, we have A: B:: B: C; then AX C-B

× B2 ÷

X B= (AX CB2), consequently, B2 AC. Substituting the

By trigonometry, R2 + tang. 2 = sec. 2. Substituting the num

99 70; then

99 98 1

70 70

70;
70

2

70; and, by Case 2, the secant No. 2

1

therefore has been added to the cosine, 70

the assumed radius; and to the assumed diameter. Now, by hy

70

pothesis, the circumference of a circle is to the diameter as 34 is to 1;

2

therefore 3 times must be added to the assumed circumference;

70

1

By Case 1, the sine of the given arc is; and, by Case 2, the tan

1

gent of the given arc is ; then, dividing the sine by the tangent, we 70 1 1 1 70

14. Therefore the sine has

$ 1

have by cancellation 570

been divided into 14 equal parts, each of which is equal to the tangent

Now, by Case 1, there were 22 sides to the given polygon, each

sis, must be added to the circumference to complete the polygon;

the circumference of the polygon No. 2.

By ARTICLE 1, the double of the secant is the constantly assumed diameter, and, by ARTICLE 2, the sum of the sines or tangents is the constantly assumed circumference.