H. N. Robinson - Elements of Geometry, Plane and Spherical Trigonometry, And Conic Sections (1854)

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ROBINSON'S CLASS BOOKS. ROBINSON'S THEORETICAL AND PRACTICAL ARITHMETIC. " KEY TO ARITHMETIC. "

ELEMENTARY TREATISE ON ALGEBRA. ALGEBRA, UNIVERSITY EDITION.

"

KEY TO ALGEBRA.

"

GEOMETRY.

"

SURVEYING AND NAVIGATION.

"

ASTRONOMY, UNIVERSITY EDITION.

"

ASTRONOMY.

"

CONCISE MATHEMATICAL OPERATIONS.

"

NATURAL PHILOSOPHY.

ROBINSON'S CLASS BOOKS.

ROBINSON'S THEORETICAL

AND PRACTICAL ARITHMETIC,

In which, in addition to the usual modes of operation, the science of numbers, the Prussian canceling system, and other important abbreviations hold a prominent place.

KEY TO ROBINSON'S ARITHMETIC. ROBINSON'S

ELEMENTARY TREATISE ON ALGEBRA,

For beginners. a

In

this work,

some Tory common subjects are presented in

new light. From the author's experience

aiid strict attention to the preparation of suitable text-books, the public ore assured that they will find this a very desirable fur the jilace it is designed to fill.

work

ROBINSON'S ALGEBRA, UNIVERSITY EDITION, Being a full course of the science in a clearer and more concise form than any other work of the kind heretofore published. It contains all the modern improvements, and develops the true spirit of the science, and is highly appreciated by tin: most important institutions of learning ia the Eastern States, as well as in the Vt'est.

KEY TO ROBINSON'S ALGEBRA, For the UPC of teachers, and for those who study without a teacher, containing, also, the Indeterminate and Diophantine analysis in concise form.

ROBINSON'S GEOMETRY, Containing Plane and Spherical Trigonometry, Conic Sections, and the necessary Logarithmic tables, for practical use. Tliis look is designed to Rive the student a knowledge of Geometry, at once theoretical, practical and efficient. The clearest methods of demonstration are employed according to the nature of the proposition, whether it IHJ strictly G'oni opposite by a ; and opposite C by let i\ii$ be a (and general form of notation) b

also

:

perpendicular by p, and we are to show that & 2 =a 2 -f-

represent the

DB by x. By

c

(th.

Also

Now 36)

^-Ka-hr) />*+

2

*

^

3

2

=c

(1) 2

(2)

B O O K

6

transposition

2

=a +c 2

This equation

Sinolium.

39

.

we have

equation (1), and subtracting (2),

By expanding

By

1

is

2

true,

D.

Q. E.

+2a#.

whatever be the value of

x,

and x may be of any value less than CD. When x is very small, B is very near D, and the line c is very near in position and value

When x=0,

to^.

c

becomes p, and the angle

and the equation becomes

right angle,

5

2

=

ABC

2

becomes a

2

-f-c

corresponding

,

to (th. 36).

THEOREM

38.

In any triangle, the square of a side opposite an acute angle is less than the square of the base, and the other side, by twice the rectangle of the base,

Lvt

and

ABC,

the distance

of

the perpendicular from the acute angle.

eith-

er figure, represent

any triangle acute

the

CB

;

C

angle,

the base, and

AD

the

perpen-

dicular, which falls

either without or

on the base.

Then we

= CB +AC 2CBX CD.

are to prove that

AB?

2

2

As in (th. 37), put AB=c, AC=b, CBa, BD=x, AD=p; and when the perpendicular falls without the base, as in the first x. figure, CDa-{-x; when it falls on the base, CD=a Considering the

first figure,

the following equations

and by the aid of

2

;,

+(a+.r)

p>+x*=c By expanding

2

=6

(1) 2 ( )

(1), and subtracting (2), c

"%.

we have

2

a 2 to both members, and transposing 2 2 2a 2 -f 2;r ) =4 2 c (

+

By

transposing

we have

2

3

a 2 +2a*=& 2

By adding

(th. 36),

:

the

c

2 ,

+a

vinculum, and resolving

it

have

r=a

3

3

-r-6

we have

2a(fi-f-#).

Q. E. D.

into

factors,

GEOMETRY.

40

we have

Considering the other figure,

(1) (2)

By By

=6

2a#

a?

subtraction

2

c

2

2 adding a to both members, and transposing 2 2 2 2 2ax c +2a -f-a

c

.

2

=6 +a

2

2

2a(a

THEOREM If in any triangle a

line be

Q. E.

a:).

of

ABC

Let

that

be a triangle, Then we are

its

base

to

prove

M.

2AM +2CM =AC +AB 2

2

AD

Draw

base, and

call

it

AB=c, CJB=2a;

DJS=a

=a-\-x, and

by

(th.

2 .

2/7

2m +2a 2

Therefore

CD

then

AM=m.

Put

+(+s) =& 2

4-2a; 2

=

2

(1) (2)

+2a 2 =6 +c +c 2

2

2

2 .

four sides of Let

Q. E.

draw

its

We

now

to

A C and BD. show,

=EC, DE=EB.

2d.

=AB

2

2

4-BC

2

1st.

That

+DC +AD

2 .

That AE AC +BD 2

bisect

sum of

be any parallelogram, and

diagonals

are

to the

=m

2

D.

40.

the parallelogram.

ABCD

But j9 2 +a; 2

.

their squares is equal

:

2

The two diagonals of any parallelogram

sum of

squares

the

THEOREM the

sum of the

36) we have the two following equations 2

middle of

AC=b, CM=a, and

x.

j

addition

to the

.

to

p*+(a-zy=--c*

By

D.

Put

MD=x ;

MB=a. Make Now

p. then

we have

2

2

perpendicular

,

this line, togetJier with twice the

square of half the side bisected, will be equal to the of the other two sides.

bisected in

2

39.

drawn from any angle

the opposite side, twice the square

c

2

each other;

the squares

of

and

all the

BOOK 41 Tlie two triangles ABE and DEQ are equal, because AB ts=DC, the angle ABE = the alternate angle EDO, and the vertical angles at E are equal therefore, AE, the side opposite the angle ABE, equal to EC, the side opposite the equal angle EDC I.

1.

;

is

:

EB, the remaining side of the one remaining side of the other triangle. also

2.

As

AD

a triangle whose base

is

have, by (th. 39),

AC

2

2

ED,

the

bisected in E,

we

equal to

2

(1)

whose base, A C, is bisected in E, we have 2 2^LE'M-2 E ,B =ABM- BC (2)

triangle

r

r2

J

J

By

is

2AE +2JSD =AD +J)C 2

As -45 C is a

A

is

adding equations (1) and (2), and observing that EB*=ED*, we have

But four times the square of the half of a line is equal to the 2 square of the whole (scholium to th. 33); therefore 4AJZ =AC*t and 4J3D 2 =DB*; and by making the

A

2

+J)

2

=AJ)

2

-}-J)

'

k

.

substitutions

&+AB*+B C*.

we have Q. E. D.

GEOMETRY

42

BOOK

II.

PROPORTION, THE word Proportion has different shades of meaning, according to the subject to which it is applied thus, when we say that a person, a building, or a vessel is well proportioned, we mean nothing :

more than that the different parts of the person or thing bear that general relation to each other which corresponds to our taste and ideas of beauty or utility, but in a more concise and geometrical sense,

Proportion is the numerical relation which one quantity bears to another of the same kind.

DEFINITIONS AND EXPLANATIONS. In Geometry, the face

to

To

a

the numerical relation which

find

another,

1st.

A solid to a solid.

3d.

surface.

between which proportion can A line to a line. 2d. A sur-

quantitities

exist, are of three kinds, only.

we must

refer

If a quantity, as

them both

to the

one quantity bears to

same standard of measure.

A

A, be contained exactly

a certain number of times

in another

n

quan-

A

is said to measure tity, B, the quantity the quantity B; and if the same quantity, A, be contained exactly a certain number is also of times in another quantity, C,

A

measure of the quantity C, and a common measure of the quanit is called B and titles C; and the quantities B and

i

1

1

i

_

(

i

(

E

said to be a

p '

'

'

'

'

C

will, evidently, bear the same relation to each other that the numbers do which represent the multiple that each quantity is of

the

common measure A.

Thus, times,

B

if

B

and

contain

C

A three

times,

and

C

contain

A

being equimultiples of the quantity

also three

A,

will

be

BOOK equal to each other tain

and

;

B

if

43

II.

contain

A

three times, and

con-

and G will be the four times, the proportion between as the proportion between the numbers 3 and 4.

same

a quantity, D, be contained as often in another quanin B, and as often in another quantity,

if

Again, tity,

G

B

A

A is contained

E, as

E

to F, or the proportion F, as A is contained in G, the ratio of between them, will be the same as the proportion between B and C; and in that case, the quantities B, C, JE, and F, are said to be a relation which is commonly expressed proportional quantities ;

B C::E:F.

thus,

:

To

find the numerical relation that any quantity, as A, has to other by quantity of the same kind as B, we simply divide any the and A, quotient may appear in the form of a fraction, thus

B

:

73

A

Now

.

this fraction, or the value of this quotient, is

always a

A

and B. numeral, whatever quantities may be expressed by To find the numerical relation between JD and E, we simply divide

E by D,

or write

-

,

which denotes the division

same quotient as when we divided

find the

;

and

if

we

JE

B

by A, then we may

write

B-D A~E If

B contains A three

we have

times,

and

just supposed, equation (

1

m (

'

D contains JE three

times, as

nothing more than saying

) is

that

3=3 When we

divide one quantity by another to find their numerical relation, the quotient thus obtained is called the ratio.

When

the ratio between two quantities is the

same as

the ratio between

two other quantities, the four quantities constitute a proportion.

On

N. B.

this single definition rests the

whole subject of geo-

metrical proportion.

On

this definition, if

A, and

we suppose

that

B is

D the same number of times E, then A to B as E to D; is

is

Or more

A

:

concisely

J3=JS:

JD.

any numbei of times

:

The

signs

:

:

meaning equal

ratio.

GEOMETRY. manifest, that if E greater than A, D will be greater

44

Now than

is

it is

A=E,

If

jB.

A relation B or ratio

then

B=D,

2

74

AD,

the expression (2 value of the chord of the half of any arc, when is

the value of

J4 C 2)* is C represents

the the

We

must, take the minus sign to C 2 as the "plus sign would give increasthe part represented by ^/4 ing, and not decreasing values.

value of the chord of the whole arc. ,

If

we

represent the chord of a given arc by C, and the chord of half

C

that arc by C,, and the chord of half that arc by t , and the chord of shall have the half that arc again by , &c., &c.,we following series 3

C

of equations

:

C= the first chord

&c.=&c.

To 60

is

equations, we

observe that in any circle the chord of apply these the to to radius (cor. equal prob. 26), and if the radius is assumed

as unity,

we

have,

C = chord of 60

(o_ vrZc'*)==C' ins. pol. of

=1.000000000

sid.

=

sid.

6 sides.

ins. pol. of

J

= chord of 30

12 sides.

.5176380902

BOOK

V.

99

Jt^C* )*=C t = chord of 15

(2

ins. pol. of

24

V"4^C|)^=C'3 =chordof

(2

ins. pol.

C*

V4

(2

of

96

chord of 3

chord of

jl^C* )*= C 6 = chord of ins. pol. of

384

768 8

9

ins. pol. of

.1308062583

sid.

45

=

.0654381655

sid.

1

52' 30'"

=

.0327234632

sid.

56' 15"

=

.0163622792

sid.

30'"= .0081812080

sid.

'"= .0040906112

sid.

28'

7"

14'

3" 45

1536 sides.

J4^CI=C = chord of

(2

=

sides.

=C = chord of ins. pol. of

30'

sides.

=C,= chord of ins. pol. of

7

192 sides.

ins. pol. of

(2

sid.

sides.

J~4^Ct )*=C5 =

(2

.2610523842

48 sides.

)*=C4 =

ins. pol. of

=

sides.

7'

=

&c.

.0020453068

sid.

3072 sides.

Hence, .0020453068X3072=6.2831814896,

is

the perimeter of an

inscribed polygon of 3072 sides when the radius is 1, or diameter 2. When the diameter is 1, the perimeter is 3.1415907498, which is a

a

little,

and but

a little, less

than the circumference, as determined by

more extended computations. Although not necessary for practical application, the following theorem for the analytical tri-section of an arc will not be

beautiful

unacceptable.

THEOREM Given, the chord of any arc,

Let

AE

to

5.

determine the chord of one third of such arc.

be the given chord, and conceive

its

arc divided into three equal parts, as represented

by

AB, BD, and DE.

Through the center draw BCG, and join ABThe two AS, CAB and ABF, are equiangular; angle FAB, being at the circumference, measured by half the arc BE, which is equal to AB, and the angle BCA, at the center, is for the

is

GEOMETRY.

100 measured by the arc angle

CBA

or

FBA,

AS; is

therefore, the angle

common

to

FAB=BCA;

but the

both triangles; therefore, the third

is equal to the third angle, AFB, angle, CAB, of the one triangle, of the other (th. 11, b. 1, cor. 2), and the two triangles are equiangular

and similar.

A

But the isosceles,

and

CBA is isosceles; therefore, the AB=AF, and we have the following CA AB AB BF :

:

Now

:

A AFB

and the proportion becomes, 1

Also,

.

.

.

:

a:

:

:

x

:

FG=2

.

also

:

:

AE=c, AB=x, CA=1. Then AF=x,

let

is

proportions

BF.

Hence

and

EF=cx,

BF=x*

x2 >

AE

As

point F,

That Or,

and

we

GB

are

two chords that intersect each other

at the

have,

GFxFB=AFy.FE

is, .

x 2)x2 =*(c

(2

.

.

.

.

a;

.

3x=

3

(th. 17, b. 3)

x)

c

AF

to be 60 degrees, then c=l, and the If we suppose the arc 3 3x= 1; a cubic equation, easily resolved by equation becomes a; Horner's method ( Robinson's Algebra, University Edition, Art. 193), giving x=. 347296-}-, the chord of 20. This again may be taken for

the value of

c,

and a second solution will give the chord of 6 many times as we please.

40',

and

so on, trisecting as

If the pupil has carefully studied the foregoing principles, he has the foundation of all geometrical knowledge; but to acquire indepen-

dence and confidence,

mind

it

is

necessary to receive such

discipline of

as the following exercises furnish.

Some

of the examples are

mere problems, some

some a combination of both.

are theorems, and

Care has been taken in their selection,

that they should be appropriate ; not very severe, not such as to try the powers of a professed geometrician, nor such as would be too trifling to engage serious attention.

EXERCISES IN GEOMETRICAL INVESTIGATION. 1

.

shall 2.

From two given meet

in the

From two

position to

draw

If

draw two equal straight

lines,

which

point, in a line given in position. given points on the same side of a line, given in two lines which shall meet in that line, and make

equal angles with 3.

points, to

same

it.

from a point without a

circle,

two

straight lines be

drawn

to

BOOK

V.

101

the concave part of the circumference, making equal angles with the line joining the same point and the center, the parts of these lines

which are intercepted within the circle, are equal. 4. If a circle be described on the radius of another

circle, any drawn from the point where they meet, to the outer circumference, is bisected by the interior one. 5. From two given points on the same side of a line given in position, to draw two straight lines which shall contain a given angle, and

straight line

be terminated in that 6. If,

line.

from any point without a

drawn touching

circle, lines "be

it,

the angle contained by the tangents is double the angle contained by the line joining the points of contact, and the diameter drawn through

one of them. 7. If,

from any two points in the circumference of a circle, there be straight lines to a point, in a tangent, to that circle, they

drawn two will

make

8.

the greatest angle

From

which

when drawn

to the point of contact.

a given point within a given circle, to draw a straight line make, with the circumference, an angle, less than any

shall

angle made by any other line drawn from that point. 9.

If

two

circles cut

each other, the greatest line that can be drawn

through the point of intersection,

is

that

which

is

parallel to the line

joining their centers. 10. If, from any point within an equilateral triangle, perpendiculars be drawn to the sides, they are, together, equal to a perpendicular

drawn from any of the angles

to the opposite side.

11. If the points of bisection of the sides of a

given triangle be

joined, the triangle, so formed, will be one-fourth of the given triangle. 12.

The

difference of the angles at the base of

any

triangle, is double

the angle contained by a line drawn from the vertex perpendicular to the base, and another bisecting the angle at the vertex. 13. If, from the three angles of a triangle, lines be drawn to the points of bisection of the opposite sides, these lines intersect each other in the same point.

14.

angle, 15.

The meet

three straight lines which bisect the three angles of a in the same point.

The two

triangles,

tri-

formed by drawing straight lines from any

point within a parallelogram to the extremities of two opposite sides, are, together, half the parallelogram. 16. The figure formed by joining the points of bisection of the sides of a trapezium, is a parallelogram. 17. If squares

be described on three sides of a right angled triangle,

GEOMETRY.

102

and the extremities of the adjacent sides be joined, the triangles so formed, are equal to the given triangle, and to each other. 18. If squares be described on the hypotenuse and sides of a right angled triangle, and the extremities of the sides of the former, and the adjacent sides of the others, be joined, the sum of the squares of the

lines joining them, will

be equal to

five

times the square of the

hypotenuse.

The

vertical angle of an oblique-angled triangle, inscribed in a greater or less than a right angle, by the angle contained between the base, and the diameter drawn from the extremity of 19.

circle, is

the base. 20. If the base of

any triangle be bisected by the diameter of

its

circumscribing circle, and, from the extremity of that diameter, a perpendicular be let fall upon the longer side, it will divide that side into

segments, one of which will be equal to half the sum, and the other to half the difference of the sides.

A

f 21. straight line drawn from the vertex of an equilateral riangle, inscribed in a circle, to any point in the opposite circumference, is

equal to the two lines together, which are drawn from the extremities of the base to the same point. 22. The straight line bisecting any angle of a triangle inscribed in a given circle, cuts the circumference in a point, which is equidistant from the extremities of the sides opposite to the bisected angle, and

from the center of a circle inscribed in the triangle.

v

a line be drawn to any point in If, the chord of an arc, the square of that line, together with the rectangle contained by the segments of the chord, will be equal to the square

from the center of a

23.

circle,

described on the radius. 24. If

two points be taken in the diameter of a

circle, equidistant

from the center, the sum of the squares of the two lines drawn from these points to any point in the circumference, will be always the same. 25.

If,

on the diameter of a semicircle, two equal

circles be described,

space included by the three circumferences, a circle be inscribed, its diameter will be f the diameter of either of the equal

and

in the

circles.

26. If

a perpendicular be drawn from the vertical angle of any

triangle to the base, the difference of the squares of the sides to the difference of the squares of the segments of the base.

i*

equal

27. The square described on the side of an equilateral triangle, is equal to three times the square of the radius of the circumscribing circle.

BOOK

103

The sum

of the sides of an isosceles triangle, is less than the of any other triangle on the same base and between the same

28.

sum

V.

parallels.

29. In

triangle, given one angle, a side adjacent to the given the difference of the other two sides, to construct the

any

angle, and triangle.

30. In any triangle, given the base, the sum of the other two sides, and the angle opposite the base, to construct the triangle.

31. In any triangle, given the base, the angle opposite to the base, and the difference of the other' two sides, to construct the triangle.

PROBLEMS REQUIRING THE AID OF ALGEBRA FOR THEIR SOLUTION. No

definite rules

can be given for the solution or construction of

the following problems ; and the pupil can have no other resources than his own natural tact, and the application of his analytical and

geometrical knowledge thus far obtained ; and if that knowledge is sound and practical, the pupil will have but little difficulty; but if his geometrical acquirements are superficial and fragmentary, the difficulties may be insurmountable hence, the ease or the difficulty which we experience in resolving such problems, is the test of an efficient or :

knowledge of theoretical geometry.

inefficient

When

a problem

is

proposed requiring the aid of Algebra, draw the

figure representing the several parts, both

unknown

known and unknown.

Rep-

known

parts by the first letters of the alphabet, and the and required parts by the final letters, &c.; and use whatever

resent, the

truths or conditions are available to obtain a sufficient

number of

equations, and the solution of such equations will give the unknown and required parts the same as in common Algebra.

But as we are unable to teach by more general precept, we give the solutions of a few examples, as a guide to the student. The first two are specimens of the most simple and easy; the last two or three are specimens of the most difficult and complex, or such as might not be readily resolved, in case solutions were not given. It might be proper to observe that different persons might draw different figures to the

more complex problems, and make

equations and give different solutions always the most simple.

PROBLEM

;

different

but the best solutions are

1.

Given, the hypotemise, and the sum of the other two sides of a right angled triangle, to determine the triangle.

GEOMETRY.

104 Let A

EC be

the

we

CB=y,ABx, A C=h,

A- Put

Then, by a given condition we

and CB-\-AB=s. have,

x+y=s And,

From

x2+3/ 2 =A 2

.

.

(th. 36, b. 1)

these two equations a solution

is

easily ob-

tained, giving,

If

A=5, and s=7,

x=4

or 3, and

y=3

or 4.

In place of putting x to represent one side, and y the other, might put (x-}-y) to represent the greater side, and (a: y} the lesser

N. B.

we

2

side; then,

.

x 2 -\-y 3

.

A ='

and 2x=s, &c.

PROBLEM

2.

Given, the base and perpendicular of a triangle, to find the side of its inscribed square.

ABC be the A- AB=b, the CDp, the perpendicular. Draw EF parallel to AB, and suppose Let

base,

equal to EG, a side of the required square; and put EF=x. it

As we

Then, by proportional

That

is,

and

:

x

:

p

EF x

:

:

:

:

have,

CD AB :

p

:

b

bp

Hence, That

CI

is, the

bx=px;

Ip

.

or,

a:

, ,

side of the inscribed square is equal to the product of the base

altitude, divided

by their sum.

PROBLEM In a

3.

about the vertical angle, and triangle, having given ine bisecting that angle and terminating in the base, to find the base. be the A> and let a circle be cirLet the sides

ABC

cumscribed about

it.

Divide the arc

AEB into

parts at the point E, and join EC. This line bisects the vertical angle (th. 9, b. 3,

two equal

scholium).

Join

Put AD=x, nnd

DEw.

BE.

DB=y, A Ca, CB=b, CD=c, The two AS, ADC and EBC,

are equiangular; from which we have, a c; or, cw-\-c*=ab w-\-c b :

:

:

:

(1)

the.

BOOK EC

But, as circle,

we

Therefore,

CD

two chords that intersect each other

are

bisects the vertical

a

:

b

...

Or,

105

.... ....

and

have,

But, as

AB

V.

:

:

cw=xy 2 xy+c =ab angle, we have,

x :y

(2)

(th. 23, b. 2)

x=f

(3)

a I z ry -\-c*=ab; or y=-Jb*-

Hence,.

in a

(th. 17, b. 3)

c*b -

c*b

And,

x and y

as

Now,

are determined, the base

Observe that equation (2)

N. B.

is

PROBLEM To angle,

is

determined.

theorem 20, book

3.

4.

determine a triangle, from the base, the line bisecting the vertical the diameter of the circumscribing circle.

and

Describe the circle on the given diameter, and divide it in two parts, in the point D,

A B,

ADxDB

shall be equal to the square so that of one half the given base.

D draw EDG EG will be the

at right angles to

Through

AB, and

given base of the

triangle.

Put

AD=n, DB=m, AB=d, DG=b.

.

Then, n-}-m=d, and nm=b 2 and these two equations will determine n and m; and therefore, n and m we shall consider as known. ;

Now, suppose EHG to be the required A> and join HIB and HA. The two 8 AHB, DBI, are equiangular, and therefore, we have,

A

>

AB HB :

But

HI

is

IB=w, we

:

:

IB

:

DB.

we will represent by c; and if we put have HB=c-\-w; then the above proportion becomes,

a given line, that

shall

d

:

c-\-w

:

Now, w can be determined by a is a known line.

:

w m :

quadratic equation;

and therefore,

IB

In the right angled

known;

therefore,

El

IG

and

are

DI

A DBI, the hypotenuse IB, and is

known.

known

(th. 36, b. 1);

and

if

base

DI

is

DB,

are

known,

GEOMETRY.

106 Lastly, let

EH=x, HG=y,

Then, by theorem 20, book

and put

El=p,

..... ...... 3,

Or v^r,

IG=q.

pq-\-c*=xy

x :y

But,

and

:

:p

(1) :

q

(th. 25, b. 2)

r x=

f9"\

{4}

And, from equations (1) and (2) we can determine x and y, the sides of the A; and thus the determination has been attained, carefully and easily, step

by

step.

PROBLEM Three equal

5.

circles touch each other externally,

of ground; what

is the

and thus

inclose one acre

diameter in rods of each of these circles

?

Draw

three equal circles to touch each other externally, and join the three centers, thus

The lines joining the forming a triangle. centers will pass through the points of contact (th. 7, b. 3).

R

Let circles

A

is

;

represent the radius of thse equal then it is obvious that each side of this

The triangle is therefore incloses the given area, and three equal sectors. each sector is a third of two right angles, the three sectors are, equal to 2.R.

equilateral,

As

and

it

together, equal to a semicircle; but the area of a semicircle,

whose

rtR'*

radius

is

R,

is

expressed by

jr-

(th. 3, b. 5, and th.

1, b. 5);

and the

rtR*

area of the whols triangle must be

A

is also

equal to

R

-

f

-{-160;

but the area of the

multiplied by the perpendicular altitude, which

.

Therefore,

Or,

.

.

RV 3= ~2~+ 160

J2 z (2,y3

rt)=320 320

3.20

=992.248

2^/33.1415926 Hence,

0.3225

.R=31.48-j- rods for the result.

PROBLEM

6.

In a right angled triangle, having given the base and the perpendicular and hypotenuse, to find these two sides.

sum

of the

BOOK

V.

PROBLEM

7.

Given, the base and altitude of a triangle, to divide

it

into three equal

parts, by lines parallel to the base.

PROBLEM In any equilateral

AJ given

from any point within,

the length

of

drawn

determine the sides.

to the three sides, to

PROBLEM In a

8. the three perpendiculars

9.

and the difference right angled triangle, having given between the hypotenuse and perpendicular (1), tofind both these two sides. the base (3),

PROBLEM In a

right angled triangle, haviny

ference between the base

and perpendicular

sides.

PROBLEM

Having

given, the area or

10.

given the hypotenuse (5),

measure of

and

the dif-

two

(1), to determine both these

11.

the space

of a rectangle inscribed

in a given triangle, to determine the sides of the rectangle.

PROBLEM In a

the ratio

having given of the base, made by a perpendicular from

triangle,

the segments

12.

of the two sides, together with both the vertical angle, to

determine the sides of the triangle.

PROBLEM In a

the base, the

having given of a line drawn from

triangle,

the length

to find the sides

13. sum of

the other

the vertical angle to the

two

and

sides,

middle of the base,

of the triangle.

PROBLEM

14.

To determine a right angled triangle; having given the lengths of two lines drawn from the acute angles to the middle of the opposite sides.

PROBLEM To determine a radius of

its

right angled triangle; having given the perimeter,

To determine a

the

triangle;

having given

16. the base, the perpendicular,

and

of the two sides.

PROBLEM To

and

inscribed circle.

PROBLEM the ratio

15.

17.

determine a right angled triangle; having given the hypotenuse, and

the side

of the inscribed square.

GEOMETRY.

108

PROBLEM To

18.

determine the radii of three equal circles, inscribed in a given

to touch each other,

and

circle,

also the circumference of the given circle.

PROBLEM

19.

In a rigid angled triangle, having given the perimeter, or sum of aU the sides, and the perpendicular let fall from the right angle on the hypotenuse, to determine the triangle; that is, its sides.

PROBLEM

20.

right angled triangle; having given the hypotenuse and of two lines, drawn from the two acute angles to the center of

To determine a the difference

the inscribed circle.

To determine a the difference

PROBLEM triangle; having given

of the two other

the rectangle, or

triangle; having

product of

from

and

the

two

22.

given the base, the perpendicular,

and

sides.

PROBLEM To

base, the perpendicular,

sides.

PROBLEM To determine a

21.

tJie

23.

determine a triangle; having given the lengths of three lines drawn the three angles to the middle of the opposite sides.

PROBLEM In a

triangle,

having given

inscribed circle.

all the

24.

three sides, to

PROBLEM

find the radius of the

25.

To

determine a right angled triangle; having given the side of the inscribed square, and the radius of the inscribed circle.

PROBLEM To

26.

determine a triangle, and the radius of the inscribed circle; having

given the lengths of three lines drawn that circle.

PROBLEM

To determine a the radius

from

the three angles to the center

27.

right angled triangle; having given the hypotenuse,

of the inscribed

circle.

of

and

BOOK

VI.

BOOK

VI

ON THE INTERSECTION OF PLANES.

DEFINITIONS, THE

14th definition of book

1,

defines a plane.

It is a superfices,

having length and breadth, but no thickness.

The

surface of

give a person

A

still

water, the side of a sheet of paper,

some idea of a

curved surface

is

may

plane.

not a plane

;

although

we sometimes

say,

" the plane of the earth's surface." 1.

If any two points be taken in a plane, and a straight

line join

the points, every point in that line is in the plane. 2. If any point in such a line should be either above or below the surface, such a surface would not be a plane. 3. straight line is perpendicular to a plane, when it makes

A

which it meets in that plane. are to each other when any straight planes perpendicular line drawn in one of the planes, perpendicular to their common right angles with every straight line 4.

Two

section, is perpendicular to the other plane.

two planes cut each other, and from any point in the line of section, two straight lines be drawn, at right angles to that line, one in the one plane, and the other in the other plane, 5.

their

If

common

the angle contained

by these two

lines is the angle

made by

the

planes. 6.

A

straight line

is

parallel to a plane

when

plane, though produced ever so far. 7. Planes are parallel to each other

though produced to any extent. 8. solid angle is one which

A

point, of

more than two plane

plane with each other.

it

does not meet the

when they do not meet,

formed by the meeting, in one angles, which are not in the same is

GEOMETRY.

110

THEOREM

1.

three straight lines meet one another, they are in one plane.

If any

BC

For conceive a plane passing through about that

to revolve

line

till it

pass through Then because the points the point E. is in and G are in that plane, the line it ; and for the same reason, the line

E

EG

EB

is

in

it

Hence

BG

and

;

the lines

is

in

it,

by

AB, CD, and

hypothesis.

BG are

all in

one plane.

Any two

Cor.

plane

straight lines

which meet each other, are

THEOREM If

a

in

one

in one plane.

and any three points whatever, are

;

2.

two planes cut one anotJier, the line

of

their

common

section is

straight line.

For

let

B

and D, any two points in the line

common

of their

be joined by the then because the points

section,

BD ;

straight line

D are both in the plane AE, the whole BD in that plane and for the same in the plane GF. The straight line BD is therefore reason BD

B

and

line

is

;

is

common common

to

both planes

;

and

it

of those Let

of intersection,

it

mil

stand at right angles to

at their point of intersection will be at right angles to

through

CD, itself.

A

any other

EF

line

and

Then

A.

drawn

in the plane, passing through

EF

t

and, of course, at right angles to the plane

(Def. 3.)

Through A, draw any

line,

A G,

of two other straight

be at right angles to the plane

lines.

AB

AB

the line of their

THEOREM.

3.

straight line stand at right angles to each

lines at their point

CD,

therefore

section.

PROPOSITION If a

is

in the plane

BOOK

EF CD,

VI.

and from any point G, draw and join FG and produce

OH

HF=AH,

AD,

parallel to

Ill

it

parallel to

D.

to

AD.

Because

Take

HG

is

we have

FH HA :

:

:

FG GD :

But, in this proportion, the first couplet is a ratio of equality therefore the last couplet is also a ratio of equality, is bisected in G. That is, GD, or the line

FG=

FD

and BF.

JowBD, EG, Now, in we have, Also, as

AFD,

the triangle

as the base

AF^+AD^=2AG +2GF 2

.

DF

theorem,

;

is

A BDF,

the base of the

FD

(1)

bisected in G, (th.

we have by

F*+D*=2BG*-{-2GF 2

.

is

2

39 b.

1.)

the same

(2)

By subtracting (1) from (2) and observing that BF* AF* =AB*, because BAF is a right angle and BD'1 AD^=AB^, because BAD is a right angle, and we shall then have, ;

AG AB*AG*=BG* This last equation shows that BAG Dividing

line

by

2,

and transposing

drawn through A,

.s

any

is

at right angles to

any

line

2 ,

and we have,

is

in the plane

a right angle.

EF, CD,

But

therefore

AG AB

in the plane, and, of course, at right

D.

angles to the plane itself.

Q. E.

PROPOSITION

PROBLEM AND THEOREM.

To draw a above

4.

straiglit line perpendicular to

Let J/2Vr be the plane, and above it. Take, DC, any plane,

and draw

From

the

A

point

A

Lastly, from gles to the line

is

the point on the

line

at right angles to

C,

draw

CB

plane, at right angles to the line

is

a plane, from a given point

it.

A, draw BC, and

AB at join

it.

on the

DC. right an-

BD.

ABC

a right angle by construction, and now if we can prove that also a right angle, then AB is at to the

last proposition.

right angles

ABD

plane, by the

GEOMETRY.

112 Because

ABC

is

a right angle,

To both members

we

have,

DC

of this equation, add

ACD

is

3

1

is

a right angle,

AC +J)C a

values in the last equation,

we

a

PROPOSITION straight lines, having the

and because latter

which shows that

ABD

have, ;

Q. E. D.

a right angle, and proves our proposition.

Two

,

and taking these

=AlP,

AEP+BD^AD* is

and we have,

Aff+(C*+D C )=A C*+D C BCD a right angle, BW+DC^BD' 2

Because

2

5.

same

THEOREM.

inclination to a plane, whether

perpendicular or oblique, are parallel to one another.

This proposition is axiomatic from our definition of parallel lines a stationary plane can have but one position, and the same in-

;

for

clination lines

;

from any fixed

position, must, of

but, for the sake of perspicuity,

we

course, give parallel

will give the following

as a demonstration.

Let

MN be a plane, and AB and

CD lines

same inclination to it. Then AB and CD are parallel.

having the

meet the plane, produce it in B and D. Join the points B and D, by the line BD, and produce it to E. The angle CDE=ABD, otherwise the two lines would not have the same inclination to the plane. But when one line, as BE, cuts two others, as AB CD, making the exterior angle, CDE, equal to If the lines do not

them

until

they do meet

the interior and opposite angle on the same side, ABE, then the and CD, are parallel. (Converse of th. 6, b. lines,

two

AB

1).

Q.

E. D.

k

PROPOSITION If two

straight lines be

6.

THEOREM.

drawn in any position through parallel

planes, they will be cut proportionally by the planes.

..

BOOK

VI.

113

Conceive three planes to be parallel, as represented in the figure, and take any points,

A

and B,

"plane at

and third planes, and

in the first

AB, which

join

passes through the second

E.

C

and Also, take any other two points, as D, in the first and third planes, and join CD, the line passing through the second plane at F. Join the two lines

by the diagonal

the second plane at G.

now

to

show

AE EB

that,

:

:

the planes are parallel,

triangles ABD and AEG,

is

:

OF FD

.

:

By comparing we have,

.

We

are

CfD= Y. EG; then, in

and

parallel

AE-.EB: X: .

A C.

the two

have, (th. 17 b. 2).

Also, as the planes are parallel,

have,

which passes through

FD

A G=X,

BD

we

OF:

:

For the sake of perspicuity, put

As

AD,

EG-, OF, and

ED,

Join

:

:

T

GF is parallel to

AC, and we

X Y :

the proportions, and applying theorem 6, book 2, . : OF: FD. O. E. D.

AE EB :

:

PROPOSITION

7.

THEOREM.

If a straight line le perpendicular to a plane, all planes passing through that line^will be perpendicular to the first-mentioned plane.

AB BC passing through AB ; this plane

Let J/"Ar be a plane, and perpenit. Let be any other

dicular to

plane, will

be perpendicular

Let

BD be

the

to

MN.

common

intersection

of the two planes, and from the point B,

draw

BE

at right angles to

Then, as

AB

is

DB.

perpendicular to the plane

MN,

it is

B

ular to every line in that plane, passing through is a But the angle therefore, right angle.

ABE

perpendic-

(def. 1, b. 6);

ABE

(def.

6,

measures the inclination of the two planes therefore, the is CB plane perpendicular to the olane MN, and thus we can show b. 6),

;

GEOMETRY.

114

AB,

that any other plane, passing through to

MN;

PROPOSITION From the same point

THEOREM.

8.

a plane, but one perpendicular can be

in

erected

the plane.

from Let and,

be perpendicular

will

D.

Q. E.

therefore, &c.

MN be

if

Ba

a plane, and let

possible,

point in

it,

BA

two perpendiculars,

and JSC, be erected. Let BD be drawn on

the plane MN, coinciding in direction with the plane passing through these two perpendiculars.

Now, line that

as a perpendicular to a plane is at right angles to every can be drawn on the plane, through the foot of the per-

pendicular, therefore,

ABD

is

CBD

a right angle, also

a right

is

angle.

ABD=

Hence, CBD; the greater equal to the less, which is must coincide with BA, and be one and absurd therefore,

BC

;

the same line

;

therefore,

from the same

PROPOSITION If two planes are perpendicular section

of

the two

D.

THEOREM.

9. to

Q. E.

point, &c.

a third plane,

the

common

inter-

planes will be perpendicular to the third plane.

BD

CB and be two planes, both perpendicular to the third plane, MN, and let be the common point to all three of the planes. Let

B

From B, draw

BA will GB,

to

BA

be

in the

this

will

erected from the is

a

common

plane

be

two

or, there

point,

section to the lines

which

may

is

PROPOSITION If a solid angle be formed by is

;

also a perpendicular

be two perpendiculars

impossible

;

therefore,

two planes BC and CD, and it .5^ and BG, on the plane MN.

therefore perpendicular to that plane.

two of them

FB

From B, draw

BD.

BA ;

same

right angles to the is

at right angles to

10. three

greater than the third.

(Prop.

3, b. 6).

BA is

at

AB

Q.E.D.

THEOREM.

plane angles, the sum of any

BOOK

VI.

BAD, DAG, BAG,

Let the three angles,

form the solid angle A. The sum of any two of these is greater than the third. When these angles are all equal, it is evident that the sum of any two is greater than the third, and the proposition needs demonstration only when one of them, as C, is greater

BA

than either of the others their

we

;

are then to prove that

less

than

line, as

BD.

it is

sum.

On

the line

take any point, B, and

AB,

the same point, B, make the angle From the point A, and on the plane

From

DC.

draw any

ABCABD,

and join

BAC, draw the angle the two plane triangles and BAE, have a common side, AB, and the angles adjacent equal (th. 14, b. 1); therefore, the two AS are, in all respects, equal; and

BAD

BAE=BAD. Now

ADAE, and BD=BE. In the triangle But,

By

BD C,

B CBAC.

PROPOSITION less

at

EAC,

opposite

and

AC

DA C,

EC.

is

op-

(Con-

DAC^EAC DAB=BAE

.

By addition,

The

therefore, the angle

).

.

.

EAC, DA=AE,

and

CD;

greater than the angle

A,

is,

But,

is

EC