Lawrence S. (Lawrence Sluter) Benson. # Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson online

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Online Library → Lawrence S. (Lawrence Sluter) Benson → Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson → online text (page 18 of 21)

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by the second corollary to the seventeenth proposition of the

sixth book to the above, we find a perfect agreement ; taking

EXERCISES. ^17

the circle as three times square of radius, we have quadrant

ACB equivalent to J AB'; hence (I. ax. 1) the semiciicle ADO

is equivalent to | AW. But (VI. 17, cor, 2) the segment AC

of the quadrant ACB is equivalent toi AB"; therefore (I. ax. 3)

we have the triangle ABC and the crescent ADC each equiva-

lent to ^ AB', thus showing the agreement between the second

corollary to the seventeenth proposition of the sixth book, and

the established truth relating to the crescent or lune.

Also, we have (VI. 17, cor. 2) the hemisphere generated by

the quadrant BNP equivalent to the solid generated by the

trapezium BSNP, and we have the solid generated by the

figure BTNP common ; thei-efore (I. ax. 3) the solid generated

by the segment BT is equivalent to the solid generated by the

figure TSN. Now, the solid generated by the segment BT is

a part of the hemisphere; hence its contents are computed by

the same radius as the hemisphere ; the solid generated by the

figure TSN is a part of the solid generated by the trapezium

BSNP ; hence its contents are also computed by the same

radius as the hemisphere. Therefore we obtain by analysis

the ti-uth, that vchen equivalent solids are generated by equiva-

lent surfaces, the generating surfaces are up n equal radii, a

truth corresponding to the truth established by the second

coroUaiy to the seventeenth proposition of the sixth book, tJiat

equivalent surfaces upon the same radius loill generate equiva-

lent solids. The synthesis then commences with the conclusion

of the analysis, and retraces its sevei-al steps, making that pre-

cede which before followed, till we arrive at the required con-

clusion. Therefore the demonstrations given in the second

corollary to the seventeenth proposition, book sixth, obtain the

conclusion from which the analyses precede. From this it ap-

pears that analysis is the instrument of investigation ; while

svnthesis affords the means of communicatino- what is already

known ; and hence, in the Elements of Euclid, the synthetic

method is followed throughout. What is now said will receive

further illustration from the solution of the following easy

problem.

Given the perimeter and angles of a triangle, to construct it.

Analysis. â€” Suppose ABC to be the required triangle, and

produce BC both ways, making BD equal to BA, and CE to

218

EXERCISES.

CA ; then DE is given, for it is equal to the sum of the throo

sides AB, BC, CA ; that is, it is equal to the given perimeter.

j^ Join AD, AE. Then (I. 1, cor.

1) the angles D and DAB are

equal, and therefore each of

them is half of ABC, because (T.

20) ABC is equal to both. The

angle D therefore is given ; and

in the same manner it may be shown that E is given, being

half of ACB. Hence the triangle ADE is given, because the

base DE, and the angles D, E are given ; and ADE being

given, ABC is also given, the angle DAB being equal to D,

and EAC equal to E.

Composition. â€” Make DE equal to the given perimeter,

the angle D equal to the half of one of the given angles, and E

equal to the half of another; draw AB, AC, maldng the angle

DAB equal to D, and EAC to E; ABC is the triangle re-

quired.

For (T. 1, cor. 2) AB is equal to BD, and AC to CE. To

these add BC, and the three, AB, BC, CA, are equal to DE,

that is, to the given perimeter. Also (I. 20) the angle ABC is

equal to D and DAB, and is therefore double of D, since D

and DAB are equal. . But D is equal to the half of one of the

given angles; therefore ABC is equal to that angle; and, in

the same manner, ACB may be proved to be equal to another

of the given angles. ABC therefore is the triangle required,

since it has its perimeter equal to the given perimeter, and its

angles equal to the given angles.

It is impossible to give rules for effecting analyses that will

answer in all cases. It may be stated, however, in a general

way, that when sums or differences are concerned, the corre-

sponding sums or differences should be exhibited in the as-

sumed figure ; that in many cases remarkable points should be

joined ; or that tlirough them lines may be drawn perpendicu-

lar or parallel to remarkable lines, or making given angles with

them ; and that circles may be described with certain radii,

and from certain points as centers; or touching certain lines,

or passing tlirough certain points. Some instances of analysis

will be given in subsequent propositions ; and the student will

EXERCISES. 219

find it useful to make analyses of many other propositions, such

as several in the Exercises.

8. A jyorism is a proposition affirming the possibility of find-

ing such conditions as will render a certain problem indeter-

minate, or capable of innumerable solutions.

Scho. 3. Porisms may be regarded as liaving their origin in

the solution of j^roblenis, which, in particular cases, on account

of pccidiar relations in their data, admit of innumerable solu-

tions ; and the proposition announcing the property or relation

which renders the problem indeterminate, is called a porism.

This will be illustrated by the solution of the following easy

problem.

Through a given point A, let it be required to draw a straight

line bisecting a given parallelogram BCDE.

Suppose AFG to be the required line, and let it cut the sides

BE, CD in F, G, and the diagonal CE in H. Then from tho

equivalent figures EBC, FBCG take FBCII, and the remaining

triangles EHF, ClIG are equal. Now, since (I. 16 and 11)

these triangles are equiangular, it is evident that they can be

equal in area only when their sides are equal ; wherefore II is

the middle point of the diagonal. The construction, therefore,

is effected by bisecting the diagonal EC in H, and drawiu"'

AFHG. For the triangles CHG, EIIF are

equiangular, and since CH, HE are equal, the

triangles are equal. To each of them add the

figure FBCH ; then the figure FBCG is equiv-

alent to the triangle EBC, that is, to half the

parallelogram BD.

Now, since the diagonal CE is given in

magnitude and position, its middle point II

is given in position, and therefore H is always a point in the

required line, wherever A is taken. Hence, so long as A and

II are different points, the straight line AHG (I. post 1) is de-

termined. This, however, is no longer so, if the given point A

be the intersection of the diagonals, that is, the point H, as in

that case only one point of the required line is known, and the

problem becomes indetermhiate, any straight line whatever,

through H, equally answering the conditions of the problem;

and we are thus led by the solution of the problem to the con-

220 EXERCISES.

elusion, that in a parallelogram a point may he found, such

that a)iy straight line wliatever drawn through it, bisects the

parallelogra7n / and this is a porism.

The seventy-sixtii pi'oposition of the Exercises, when con-

sidered in a particular manner, affords another instance of a

porism ; as it appears that if a circle and a point D or E be

given, another point E or D may be found, such that any circle

whatever, desciibed through D and E, will bisect the circum-

ference of the given circle; and this may be regarded as the

indeterminate case of the problem, in which it is required,

through two given points, to describe a circle bisecting tlie cir-

cumference of another given circle, â€” a problem which is always

determinate, except when the points are situated in the man-

ner supposed in the proposition.

9. Isopjeritnetrical figures are such as have their perimeters,

or bounding lines, equal.

10, The general problem of the tangencies, as understood by

the ancients, is as follows: Of three points, three straight lines,

and three circles of given radii, any three being given in posi-

tion ; it is required to describe a circle passing through the

points, and touching the straight lines and circles. This gen-

eral problem comprehends ten subordinate ones, the data of

which are as follows: (l.) three points; (2.) two points and a

straight line ; (3.) two points and a circle ; (4.) a point and two

straight lines ; (5.) a point, a straight line, and a circle ; (G.) a

point and two circles; (7.) three straight lines; (8.) two

straight lines and a circle ; (9.) a straight line and two circles;

and (10.) three circles. The first and seventh of these are the

second and fifth corollaries of the twenty-fifth proposition of

the third book.

If a circle be continually diminished, it may be regarded as

becoming ultimately a point. By being continually enlarged,

on the contrary, it may have its curvature so much diminished

that any portion of its circumference may be made to differ in

as small a degree as we please from a straight line. Viewing

the subject in this light, we may regard the first nine of the

problems now mentioned, as comprehended in the tenth. Thus,

we shall have the first, by supposing the circles to become in-

finitely small ; the seventh, by supposing them infinitely great;

EXEHCISES.

221

the fifth, by taking one of them infinitely small, one infinitely

great, and one as a circle of finite magnitude; and so on with

regard to the othei'S. These views of tlie subject tend to illus-

trate it ; but they do not assist in the solution of the problems.

Hcho. 4. In tiie fifth problem, the straight line may fall with-

out the circle, may cut it, or may touch it ; the point may be

without the circle, withfn it, or in its circumference ; or it may

be in the given straight line, or on either side of it ; and it will

be an interesting exercise for the student, in this and many

other problems, to consider the variations arising in the solu-

tion from such changes in the relations of the data, and to de-

termine what relations make the solution possible, and what

render it impossible. It may also be remarked, that in many

problems there will be slight variations in the proofs of differ-

ent solutions of the same problem, even when there is no

chi"ige in the method of solution; such as in the present in-

stance, when the required circle is touched externally, and

Avhen internally. Thus, while in one case angles may coincide,

in another the corresponding ones may be vertically ojiposite;

and the reference luay sometimes be to the coJiverse of the first

corollary and sometimes to the converse of the second corollary

of the eighteenth proposition, book third. It is, in general, un-

necessary to point out these variations, as, though they merit

the attention of the student, they occasion no difficulty.

PROPOSITIONS.

Prop. I. â€” Tiieor. â€” If an angle of a triangle, he bisected by a

straight line^ which likewise cuts the base,, the rectangle con-

tained by the sides of the triangle is equivalent to the rectangle

contained by the segments of the base,, together with the square

of the straight line bisecting the angle.

Let ABC be a triangle, and let the angle BAC be bisected

by AD; the rectangle BA.AC is equal to the rectangle BD.DC,

together with the square of AD.

Describe the circle (III. 25, cor.) ACB about the triangle ;

produce AD to meet the circumference in E, and join EC.

Then (hyp.) the angles BAD, CAE are equal ; as are also (IIL

18, cor. 1) the angle B and E, for they are in the same segment;

222 EXERCISES.

thej-efore (V. 3, cor.) in the ti-iangles ABD, AEC, as BA : AD

:: EA : AC; and consequently (V. 10, cor.) the rectangle BA.

AC is equivalent to EA.AD, that is (II. 3), ED.DA, together

with the square of AD. But (III. 20) the rectangle ED.DA ia

equivalent to the rectangle BD.DC ; therefore the rectangle

BA.AC is equivalent to BD.DC, together with the square of

AD ; wherefore, if an angle, etc.

Scho. From this proposition, in connection -with the fourth

proposition of fifth book, we have the means of computing AD,

when the sides are given in numbers. For, by the fourth, BA:

AC :: BD : DC; and, by composition, BA+AC : AC : :

BC : DC. This analogy gives DC, and BD is then found by

taking DC from BC. But by this proposition BA. AC = BD.DC

+ AD-; therefore from BA.AC take BD.DC, and the square

root of the remainder will be AD,

In a similar manner, from the fourth proposition of the fifth

book, and the second proposition of these Exercises, the line

bisecting the exterior vertical angle may be computed; and

from the third proposition of these Exercises, in connection

with the eleventh or twelfth of the second book, the diameter

of the circumscribed circle may be computed, when the sides

of the triangle are given in numbers.

Prop. II. â€” Theor. â€” If an exterior angle of a triangle be

bisected by a straight line, which cuts tlie base produced ;

the rectangle contained by the sides of the triangle, and the

square of the bisecting line are together equivalent to the rect-

angle contained by the segments of the base intercepted between

its extremities and the bisecting line.

Let, in the foregoing diagram, the exterior angle ACF of the

triangle BAC be bisected by HC ; the rectangle BC. AC and

EXERCISES. 223

the square of HC are together equivalent to the rectangle

BH.AII.

Describe the circle (III. 25, cor.) ABEC about the triangle

BAG; produce HC (I. post. 2) to E, and join EA. Then,

since (l>yp.) the angles FCII and HCA are equal, their supple-

ments FCE and ACE (T. def. 20 and ax. 3) are also equal ; and

(III. 18, cor. 1) B and E are equal. Therefore (V. 3, cor.) in

the triangles BCH and EAC, BC : CH : : EC : AC, and con-

sequently (V. 10, cor.) the rectangle BC. AC is equivalent to

EC.CII. To each add square of CII; therefore BC.AC+CIP

are equivalent to EC.CH+Cff; or (II. 3) BC.AC + CIP are

equivalent to EH.CH ; or (III. 21, cor.) BC.AC + CIP are

equivalent to BH. AH. Therefore, if an exterior angle, etc.

Prop. III. â€” Theor. â€” If from an angle of a triangle a per-

pendicular be drawn to the basej the rectangle contained by

the sides of the triangle is equivalejit to the rectangle contained

by the perpendicular and iAe diameter of the circle described

about the triangle.

Also, in the foregoing diagram, let ABC be a triangle, AL

the perpendicular from the angle A to BC ; and AE a diameter

of the circumscribed circle ABEC; the rectangle BA.AC is

equivalent to the rectangle AL.AE.

Join EC. Then the right angle BLA is equal (III. 11) to the

angle EC A in a semicircle, and (HI. 18, cor. 1) the angle B to

the angle E in the same segment ; therefore (V. 3, cor.) as

BA : AL : : EA : AC; and consequently (V. 10, cor.) the

rectangle BA.AC is equivalent to the rectangle EA.AL. If,

therefore, from an angle of a triangle, etc.

Prop. IV. â€” Theor. â€” The rectam^le contained by the diago-

nals of a quadrilateral inscribed in a circle, is equivalent to

both the rectangles contained by its opposite sides.

Let ABCD be a quadrilateral inscribed in a circle, and join

AC, BD ; the rectangle AC.BD is equivalent to the two rect-

angles AB.CD and AD.BC.

Make the angle ABE equal to DBC, and take each of them

from the whole angle ABC ; then the remaining angles CBE,

ABD are equal; and (HI. 18, cor. l) the angles ECB, ADB

are equal. Therefore (V. 3, cor.) in the triangles ABD, EBC,

224

EXERCISES.

as BC : CE

BD.CE

Again

BD : DA; wlience (V. 10, cor.) BC.DA=

in the triangles BAE, BDC, because (const.)

the angles ABE, DBC are equal, as also

(III. 18, cor. 1) BAE, BDC; therefore

(V. 3, cor.) as BA : AE : : BD : DC;

^\ hence (V. 10, cor.) BA.DC==BD.AE.

' Add these equivalent rectangles to the

equivalents BC.DA and BD.CE; then

BA.DC + BC.DA = BD.CE + BD.AE, or

(II. 1) BA.DC + BC.DA=BD.AC. There-

fore, the rectangle, etc.

Cor. 1. If the sides AD, DC, and consequentlj^ (TIT. 16, cor.

1) the arcs AD, DC, and the angles ABD, CBD be equal, the

rectangle BD.AC is equivalent to AB.AD together with BC. AD,

or (II. 1) to the rectangle under AD, and the sum of ABand BC.

Hence (V. 10, cor.) AB-f-BC : BD : : AC : AD or DC.

Cor. 2. If AC, AD, CD be all equal, the last analogy be-

comes AB+BC : BD : : AD : AD ; whence AB + BC=BD.

Hence in an equilateral triangle inscribed in a circle, a straight

line drawn from the vertex to a point in the arc cut oiF by the

base is equal to the sum of the chords drawn from that point

to the extremities of the base.

Prop. V. â€” Theor. â€” The diagonals of a qtiadrilateral in-

scribed in a circle, are proportional to the sums of the rectan-

gles contained by the sides meet'nr/ at their extremities.

Let ABCD be a quadi-ilateral inscribed in a circle, and AC,

BD its diagonals; AC : BD : : BA.AD-h

BC.CD : AB.BC+AD.DC.

If AC, BD cut one another perpendicu-

larly in L, then (Ex. 2) AK being the diame-

ter of the circle, AL. AK = BA.AD, and CL.

AK=:BC.CD; whence, by addition (I. ax.

2), AL.AK + CL.AK, or (II. 1) AC.AKz=

B.\.AD+BC.CD. In a similar manner,

it would be proved that BD.AK = AB.BC

Hence ACIAK : BD.AK, or (V. 1) AC : BD : â€¢

BA.AD + BC.CD : AB.BC + AD.DC.

But if AC be not perpendicular to BD, draw AEF perpen-

+ AD.DC.

EXERCISES.

225

dicnlar and CF parallel to BD, and DGH perpendicular and

BH parallel to AC. Then, because EF is equal to the perpen-

dicular drawn from C to BI), and GH equal to the one drawn

from B to AC ; it would be prov-

ed as before, that AP\AIv = BA.

AD + BC.CD, and I)n.AK =

AB.BC + AD.DC. Hence, AF.

AK : DH.AK, or (V. l) AF :

DII :: BA.AD-^BC.CD : AB.

BC + AD.DC. But the triani^les

AFC, DIIB are equiangular, liav-

in<x the rio;ht anojes Fand H, and

the angles ACF, DBH, each equal

(I. 16) to ALD; therefore (V. 3)

AF : AC : : DH : DB, and alternately AF : T3IT : : AC :

DB. Hence (IV. 7) the foregoing analogy becomes AC : BD

: : BA.AD + BC.CD : AB.BC-hAD.DC. Wherefore, the

diagonals, etc.

Scho. From this proposition and the last, when the sides of

a quadrilateral inscribed in a circle are given, we can find the

ratio of the diagonals and their rectangle, and thence (V, 15)

the diagonals themselves. Also, if the sides be given in num-

bers, we can compute the diagonals. Thus, let the sides taken

in succession round the figure be 50, 78, 104, and 120. Then,

the ratio of the diagonals will be that of 50 x 784-104 x 120 to

60x120-1-78x104; that is, 16880 to 14112, or 65 to 56, by-

dividing by 252. Again, the rectangle of the diagonals is 50 x

104-1-78x120, or 14560. But similar rectilineal figures are as

the squares of the corresponding sides, and consequently the

sides are as the square roots of the areas ; therefore, taking 65

and 56 as the sides of a rectangle, we have its area equal to

3640; and ^ 3640 is to 4/14560, or -^Z 3640 is to |/ (4x3640),

that is, 1 : 2 : : 65 : 130 : : 56

112 ; so that 130 and 112

are the diagonals.

Prop. VI. â€” Theor. â€” If in a straight line drawn through

the center of a circle^ and on the same side of (he center, (wo

povits be taken so that the radius is a mean proportional her-

tween their distances from, the cei^ter / two straight lines drawn

15

226 EXEECISES.

from, those points to any point whatever in the circumference^

are proportional to the segments into which the circumference

divides the straight line intercepted leticeen the same points.

Let ABC l)e a circle, and CAE a straight line drawn throngh

its center D ; if ED : DA : : DA : DF, and if BE, BF be

drawn from any point B of the circumference ; EB : BF : :

EA : AF.

Join AB, BD, Then, since DB is equal to DA, we have

(hyp.) ED : DB :: DB : DF.

The two triangles EDB, BDF,

therefore, have their sides about

the common angle D proportion-

al ; wherefore (V. 6) the angle

E is equal to FBD. Now the

angle liAD is equal (I. 20) to the

two angles E and EBA, and also

(I. 1, cor.) to ABD ; wherefore E and EBA are (I. ax. 1) equal

to ABD. From these take the equal angles E and FBD, and

(I. ax. 3) the remaining angles EBA, ABF are equal ; and

therefore (V. 4) in the triangle EBF, EB : BF : : EA : AF.

If, therefore, in a straight line, etc.

Cor. Join BC, and produce EB to G. Then, since ABC is

(III. 11) a right angle, it is equal to the two EBA, CBG.

From these equals take the equal angles ABF, EBA, and the

remainders FBC, CBG are equal ; and therefore (V. 4, 2d case)

EB : BF : : EC : CF. But it has been proved that EB :

BF : : EA : AF ; therefore (IV. 1) EA : AF : : EC : CF.

Hence, if the segments EA, AF be given, the point C may be

determined by the method shown in the third corollary to the

sixteenth proposition of the first book; and the circle ABC

may then be described, its diameter AC being determined.

Scho. The circle may also be determined in the following

manner: Since (hyp.) ED : DA : : DA : DF, by division,

EA : DA : : AF : DF ; whence, alternately and by division,

EAâ€” AF : AF : : AF : DF. Hence DF is a third proper-

tional to the difference of E A, AF, and to AF, the less ; and

thus the center D is determined. From the last analogy also

we obtained (IV. 11) EAâ€” AF : EA : : AF : AD; an anal-

ogy which serves the same purpose, since it shows that the ra-

EXERCISES. 227

dius of the circle is a fourth proportional to the difference of

the segments EA, AF, and to those segments themselves.

Prob. VII, â€” TnEOR. â€” The perpendiculars drawn from the

three angles of any triangle to the ojjposite sides, intersect one

another in the same point.

If the triangle be right-angled, it is plain that all the perpen-

diculars pass through the right angle. But if it be not right-

angled, let ABC be the triangle, and about it describe a circhi;

then, B and C being acute angles, draw ADE perpendicular to

BC, cutting BC in D, and the circumference in E; and make

DF equal to DE ; join BF and produce it, if necessary, to cut

AC, or AC produced in G ; BG is perpeixlicular to AC. Join

BE ; and because FD is equal to DE, the angles at D right

angles, and DB common to the two triangles FDB, EDB, the

angle FBD is equal (I. 3) to EBD ; but

(III. 18, cor. 1) CAD, EBD are also equal,

because they are in the same segment ;

therefore CAD is equal to FBD or GBC,

But the angle ACB is common to the tv o

triangles ACD, BCG; and therefore (I,

20, cor. 5) the remaining angles ADC,

BGC are equal ; but (const.) ADC is a

right angle; therefore also BGC is a right angle, and BG is

perpendicular to AC, In the same manner it would be shown

that a straight line CII, drawn through C and F, is perpendicu-

lar to AB. The three perpendiculars therefore all pass through

F ; wherefore, the perpendiculars, etc,

Scho. This limitation prevents the necessity of a different

ease, which would arise if the perpendicular AD fell without

the triangle. If the angle A be obtuse, the point F lies with-

out the circle, and BF, not produced, cutsAC produced. The

proof, however, is the same, and it is very easy and obvious.

Another easy and elegant proof, of which the following is an

outline, is given in Garnier's "Reciproques," etc., Theor. III.,

page 78 : Draw BG and CH perpendicular to AC and AB ;

join GH, and about the quadrilaterals AHFG and BHGC de-

8cribe circles, which can be done, as is easily shown ; draw

also AFD. Then the angles BAD, BCH are equal, each of

223 EXERCISKS.

them being equal (TIT, 18, cor. 1) to HGF ; and the angle ABO

"being coniinoii, ADB is equal (I. 20) to BHC, and is therefore

a right angle.

Prop. VTTT. â€” Theor. â€” From AB, the greater side of the tri-

angle ABC, cut off AD equal to AC, and join DC; draw AE

bisecting the vertical angle BAC, andjoiji DE ; d aw also AF

perpendicular to BC, and DG parallel to AE. Then (1.) the

angle DEB is eqtdvalent to the difference of the angles at the

base, ACB, ABC ; or of BAF, CAF, or of AEB, AEC ; and

DE is equal to EC ; (2.) the angles BCD, EAF are each equiv-

alent to half the same difference ; (3.) ADC or ACD is equiva-

lent to half the sum of the angles at the base, or to the comple-

ment of half the vertical angle ; (4.) BG is equivalent to the

difference of the segments BE, EC, made by the line bisecting

the vertical angle. ^

1. In the triangles AED, AEC, AD, AC are equal, AE com-

mon, and the contained angles equal ; therefore (I. 3) DE is

equal to EC, the angle ADE to ACE, and AED to AEC. But

(I. 20) because BD is produced, the

angle ADE is equivalent to B and

BED; therefore BED is the differ-

ence of B and ADE, or of B and

ACB. Also BED is the difference

of AEB, AED, or of AEB, AEC.

Again : ABF, BAF are equivalent

to ACF, CAF, each pair being (L

20, cor. 3) equivalent to a right angle. Take away ABF;

then, because the difference of ACF, ABF is BED, there re-

mains BAF equivalent to BED, CAF ; that is, BED is the

difference of BAI\ CAF.

sixth book to the above, we find a perfect agreement ; taking

EXERCISES. ^17

the circle as three times square of radius, we have quadrant

ACB equivalent to J AB'; hence (I. ax. 1) the semiciicle ADO

is equivalent to | AW. But (VI. 17, cor, 2) the segment AC

of the quadrant ACB is equivalent toi AB"; therefore (I. ax. 3)

we have the triangle ABC and the crescent ADC each equiva-

lent to ^ AB', thus showing the agreement between the second

corollary to the seventeenth proposition of the sixth book, and

the established truth relating to the crescent or lune.

Also, we have (VI. 17, cor. 2) the hemisphere generated by

the quadrant BNP equivalent to the solid generated by the

trapezium BSNP, and we have the solid generated by the

figure BTNP common ; thei-efore (I. ax. 3) the solid generated

by the segment BT is equivalent to the solid generated by the

figure TSN. Now, the solid generated by the segment BT is

a part of the hemisphere; hence its contents are computed by

the same radius as the hemisphere ; the solid generated by the

figure TSN is a part of the solid generated by the trapezium

BSNP ; hence its contents are also computed by the same

radius as the hemisphere. Therefore we obtain by analysis

the ti-uth, that vchen equivalent solids are generated by equiva-

lent surfaces, the generating surfaces are up n equal radii, a

truth corresponding to the truth established by the second

coroUaiy to the seventeenth proposition of the sixth book, tJiat

equivalent surfaces upon the same radius loill generate equiva-

lent solids. The synthesis then commences with the conclusion

of the analysis, and retraces its sevei-al steps, making that pre-

cede which before followed, till we arrive at the required con-

clusion. Therefore the demonstrations given in the second

corollary to the seventeenth proposition, book sixth, obtain the

conclusion from which the analyses precede. From this it ap-

pears that analysis is the instrument of investigation ; while

svnthesis affords the means of communicatino- what is already

known ; and hence, in the Elements of Euclid, the synthetic

method is followed throughout. What is now said will receive

further illustration from the solution of the following easy

problem.

Given the perimeter and angles of a triangle, to construct it.

Analysis. â€” Suppose ABC to be the required triangle, and

produce BC both ways, making BD equal to BA, and CE to

218

EXERCISES.

CA ; then DE is given, for it is equal to the sum of the throo

sides AB, BC, CA ; that is, it is equal to the given perimeter.

j^ Join AD, AE. Then (I. 1, cor.

1) the angles D and DAB are

equal, and therefore each of

them is half of ABC, because (T.

20) ABC is equal to both. The

angle D therefore is given ; and

in the same manner it may be shown that E is given, being

half of ACB. Hence the triangle ADE is given, because the

base DE, and the angles D, E are given ; and ADE being

given, ABC is also given, the angle DAB being equal to D,

and EAC equal to E.

Composition. â€” Make DE equal to the given perimeter,

the angle D equal to the half of one of the given angles, and E

equal to the half of another; draw AB, AC, maldng the angle

DAB equal to D, and EAC to E; ABC is the triangle re-

quired.

For (T. 1, cor. 2) AB is equal to BD, and AC to CE. To

these add BC, and the three, AB, BC, CA, are equal to DE,

that is, to the given perimeter. Also (I. 20) the angle ABC is

equal to D and DAB, and is therefore double of D, since D

and DAB are equal. . But D is equal to the half of one of the

given angles; therefore ABC is equal to that angle; and, in

the same manner, ACB may be proved to be equal to another

of the given angles. ABC therefore is the triangle required,

since it has its perimeter equal to the given perimeter, and its

angles equal to the given angles.

It is impossible to give rules for effecting analyses that will

answer in all cases. It may be stated, however, in a general

way, that when sums or differences are concerned, the corre-

sponding sums or differences should be exhibited in the as-

sumed figure ; that in many cases remarkable points should be

joined ; or that tlirough them lines may be drawn perpendicu-

lar or parallel to remarkable lines, or making given angles with

them ; and that circles may be described with certain radii,

and from certain points as centers; or touching certain lines,

or passing tlirough certain points. Some instances of analysis

will be given in subsequent propositions ; and the student will

EXERCISES. 219

find it useful to make analyses of many other propositions, such

as several in the Exercises.

8. A jyorism is a proposition affirming the possibility of find-

ing such conditions as will render a certain problem indeter-

minate, or capable of innumerable solutions.

Scho. 3. Porisms may be regarded as liaving their origin in

the solution of j^roblenis, which, in particular cases, on account

of pccidiar relations in their data, admit of innumerable solu-

tions ; and the proposition announcing the property or relation

which renders the problem indeterminate, is called a porism.

This will be illustrated by the solution of the following easy

problem.

Through a given point A, let it be required to draw a straight

line bisecting a given parallelogram BCDE.

Suppose AFG to be the required line, and let it cut the sides

BE, CD in F, G, and the diagonal CE in H. Then from tho

equivalent figures EBC, FBCG take FBCII, and the remaining

triangles EHF, ClIG are equal. Now, since (I. 16 and 11)

these triangles are equiangular, it is evident that they can be

equal in area only when their sides are equal ; wherefore II is

the middle point of the diagonal. The construction, therefore,

is effected by bisecting the diagonal EC in H, and drawiu"'

AFHG. For the triangles CHG, EIIF are

equiangular, and since CH, HE are equal, the

triangles are equal. To each of them add the

figure FBCH ; then the figure FBCG is equiv-

alent to the triangle EBC, that is, to half the

parallelogram BD.

Now, since the diagonal CE is given in

magnitude and position, its middle point II

is given in position, and therefore H is always a point in the

required line, wherever A is taken. Hence, so long as A and

II are different points, the straight line AHG (I. post 1) is de-

termined. This, however, is no longer so, if the given point A

be the intersection of the diagonals, that is, the point H, as in

that case only one point of the required line is known, and the

problem becomes indetermhiate, any straight line whatever,

through H, equally answering the conditions of the problem;

and we are thus led by the solution of the problem to the con-

220 EXERCISES.

elusion, that in a parallelogram a point may he found, such

that a)iy straight line wliatever drawn through it, bisects the

parallelogra7n / and this is a porism.

The seventy-sixtii pi'oposition of the Exercises, when con-

sidered in a particular manner, affords another instance of a

porism ; as it appears that if a circle and a point D or E be

given, another point E or D may be found, such that any circle

whatever, desciibed through D and E, will bisect the circum-

ference of the given circle; and this may be regarded as the

indeterminate case of the problem, in which it is required,

through two given points, to describe a circle bisecting tlie cir-

cumference of another given circle, â€” a problem which is always

determinate, except when the points are situated in the man-

ner supposed in the proposition.

9. Isopjeritnetrical figures are such as have their perimeters,

or bounding lines, equal.

10, The general problem of the tangencies, as understood by

the ancients, is as follows: Of three points, three straight lines,

and three circles of given radii, any three being given in posi-

tion ; it is required to describe a circle passing through the

points, and touching the straight lines and circles. This gen-

eral problem comprehends ten subordinate ones, the data of

which are as follows: (l.) three points; (2.) two points and a

straight line ; (3.) two points and a circle ; (4.) a point and two

straight lines ; (5.) a point, a straight line, and a circle ; (G.) a

point and two circles; (7.) three straight lines; (8.) two

straight lines and a circle ; (9.) a straight line and two circles;

and (10.) three circles. The first and seventh of these are the

second and fifth corollaries of the twenty-fifth proposition of

the third book.

If a circle be continually diminished, it may be regarded as

becoming ultimately a point. By being continually enlarged,

on the contrary, it may have its curvature so much diminished

that any portion of its circumference may be made to differ in

as small a degree as we please from a straight line. Viewing

the subject in this light, we may regard the first nine of the

problems now mentioned, as comprehended in the tenth. Thus,

we shall have the first, by supposing the circles to become in-

finitely small ; the seventh, by supposing them infinitely great;

EXEHCISES.

221

the fifth, by taking one of them infinitely small, one infinitely

great, and one as a circle of finite magnitude; and so on with

regard to the othei'S. These views of tlie subject tend to illus-

trate it ; but they do not assist in the solution of the problems.

Hcho. 4. In tiie fifth problem, the straight line may fall with-

out the circle, may cut it, or may touch it ; the point may be

without the circle, withfn it, or in its circumference ; or it may

be in the given straight line, or on either side of it ; and it will

be an interesting exercise for the student, in this and many

other problems, to consider the variations arising in the solu-

tion from such changes in the relations of the data, and to de-

termine what relations make the solution possible, and what

render it impossible. It may also be remarked, that in many

problems there will be slight variations in the proofs of differ-

ent solutions of the same problem, even when there is no

chi"ige in the method of solution; such as in the present in-

stance, when the required circle is touched externally, and

Avhen internally. Thus, while in one case angles may coincide,

in another the corresponding ones may be vertically ojiposite;

and the reference luay sometimes be to the coJiverse of the first

corollary and sometimes to the converse of the second corollary

of the eighteenth proposition, book third. It is, in general, un-

necessary to point out these variations, as, though they merit

the attention of the student, they occasion no difficulty.

PROPOSITIONS.

Prop. I. â€” Tiieor. â€” If an angle of a triangle, he bisected by a

straight line^ which likewise cuts the base,, the rectangle con-

tained by the sides of the triangle is equivalent to the rectangle

contained by the segments of the base,, together with the square

of the straight line bisecting the angle.

Let ABC be a triangle, and let the angle BAC be bisected

by AD; the rectangle BA.AC is equal to the rectangle BD.DC,

together with the square of AD.

Describe the circle (III. 25, cor.) ACB about the triangle ;

produce AD to meet the circumference in E, and join EC.

Then (hyp.) the angles BAD, CAE are equal ; as are also (IIL

18, cor. 1) the angle B and E, for they are in the same segment;

222 EXERCISES.

thej-efore (V. 3, cor.) in the ti-iangles ABD, AEC, as BA : AD

:: EA : AC; and consequently (V. 10, cor.) the rectangle BA.

AC is equivalent to EA.AD, that is (II. 3), ED.DA, together

with the square of AD. But (III. 20) the rectangle ED.DA ia

equivalent to the rectangle BD.DC ; therefore the rectangle

BA.AC is equivalent to BD.DC, together with the square of

AD ; wherefore, if an angle, etc.

Scho. From this proposition, in connection -with the fourth

proposition of fifth book, we have the means of computing AD,

when the sides are given in numbers. For, by the fourth, BA:

AC :: BD : DC; and, by composition, BA+AC : AC : :

BC : DC. This analogy gives DC, and BD is then found by

taking DC from BC. But by this proposition BA. AC = BD.DC

+ AD-; therefore from BA.AC take BD.DC, and the square

root of the remainder will be AD,

In a similar manner, from the fourth proposition of the fifth

book, and the second proposition of these Exercises, the line

bisecting the exterior vertical angle may be computed; and

from the third proposition of these Exercises, in connection

with the eleventh or twelfth of the second book, the diameter

of the circumscribed circle may be computed, when the sides

of the triangle are given in numbers.

Prop. II. â€” Theor. â€” If an exterior angle of a triangle be

bisected by a straight line, which cuts tlie base produced ;

the rectangle contained by the sides of the triangle, and the

square of the bisecting line are together equivalent to the rect-

angle contained by the segments of the base intercepted between

its extremities and the bisecting line.

Let, in the foregoing diagram, the exterior angle ACF of the

triangle BAC be bisected by HC ; the rectangle BC. AC and

EXERCISES. 223

the square of HC are together equivalent to the rectangle

BH.AII.

Describe the circle (III. 25, cor.) ABEC about the triangle

BAG; produce HC (I. post. 2) to E, and join EA. Then,

since (l>yp.) the angles FCII and HCA are equal, their supple-

ments FCE and ACE (T. def. 20 and ax. 3) are also equal ; and

(III. 18, cor. 1) B and E are equal. Therefore (V. 3, cor.) in

the triangles BCH and EAC, BC : CH : : EC : AC, and con-

sequently (V. 10, cor.) the rectangle BC. AC is equivalent to

EC.CII. To each add square of CII; therefore BC.AC+CIP

are equivalent to EC.CH+Cff; or (II. 3) BC.AC + CIP are

equivalent to EH.CH ; or (III. 21, cor.) BC.AC + CIP are

equivalent to BH. AH. Therefore, if an exterior angle, etc.

Prop. III. â€” Theor. â€” If from an angle of a triangle a per-

pendicular be drawn to the basej the rectangle contained by

the sides of the triangle is equivalejit to the rectangle contained

by the perpendicular and iAe diameter of the circle described

about the triangle.

Also, in the foregoing diagram, let ABC be a triangle, AL

the perpendicular from the angle A to BC ; and AE a diameter

of the circumscribed circle ABEC; the rectangle BA.AC is

equivalent to the rectangle AL.AE.

Join EC. Then the right angle BLA is equal (III. 11) to the

angle EC A in a semicircle, and (HI. 18, cor. 1) the angle B to

the angle E in the same segment ; therefore (V. 3, cor.) as

BA : AL : : EA : AC; and consequently (V. 10, cor.) the

rectangle BA.AC is equivalent to the rectangle EA.AL. If,

therefore, from an angle of a triangle, etc.

Prop. IV. â€” Theor. â€” The rectam^le contained by the diago-

nals of a quadrilateral inscribed in a circle, is equivalent to

both the rectangles contained by its opposite sides.

Let ABCD be a quadrilateral inscribed in a circle, and join

AC, BD ; the rectangle AC.BD is equivalent to the two rect-

angles AB.CD and AD.BC.

Make the angle ABE equal to DBC, and take each of them

from the whole angle ABC ; then the remaining angles CBE,

ABD are equal; and (HI. 18, cor. l) the angles ECB, ADB

are equal. Therefore (V. 3, cor.) in the triangles ABD, EBC,

224

EXERCISES.

as BC : CE

BD.CE

Again

BD : DA; wlience (V. 10, cor.) BC.DA=

in the triangles BAE, BDC, because (const.)

the angles ABE, DBC are equal, as also

(III. 18, cor. 1) BAE, BDC; therefore

(V. 3, cor.) as BA : AE : : BD : DC;

^\ hence (V. 10, cor.) BA.DC==BD.AE.

' Add these equivalent rectangles to the

equivalents BC.DA and BD.CE; then

BA.DC + BC.DA = BD.CE + BD.AE, or

(II. 1) BA.DC + BC.DA=BD.AC. There-

fore, the rectangle, etc.

Cor. 1. If the sides AD, DC, and consequentlj^ (TIT. 16, cor.

1) the arcs AD, DC, and the angles ABD, CBD be equal, the

rectangle BD.AC is equivalent to AB.AD together with BC. AD,

or (II. 1) to the rectangle under AD, and the sum of ABand BC.

Hence (V. 10, cor.) AB-f-BC : BD : : AC : AD or DC.

Cor. 2. If AC, AD, CD be all equal, the last analogy be-

comes AB+BC : BD : : AD : AD ; whence AB + BC=BD.

Hence in an equilateral triangle inscribed in a circle, a straight

line drawn from the vertex to a point in the arc cut oiF by the

base is equal to the sum of the chords drawn from that point

to the extremities of the base.

Prop. V. â€” Theor. â€” The diagonals of a qtiadrilateral in-

scribed in a circle, are proportional to the sums of the rectan-

gles contained by the sides meet'nr/ at their extremities.

Let ABCD be a quadi-ilateral inscribed in a circle, and AC,

BD its diagonals; AC : BD : : BA.AD-h

BC.CD : AB.BC+AD.DC.

If AC, BD cut one another perpendicu-

larly in L, then (Ex. 2) AK being the diame-

ter of the circle, AL. AK = BA.AD, and CL.

AK=:BC.CD; whence, by addition (I. ax.

2), AL.AK + CL.AK, or (II. 1) AC.AKz=

B.\.AD+BC.CD. In a similar manner,

it would be proved that BD.AK = AB.BC

Hence ACIAK : BD.AK, or (V. 1) AC : BD : â€¢

BA.AD + BC.CD : AB.BC + AD.DC.

But if AC be not perpendicular to BD, draw AEF perpen-

+ AD.DC.

EXERCISES.

225

dicnlar and CF parallel to BD, and DGH perpendicular and

BH parallel to AC. Then, because EF is equal to the perpen-

dicular drawn from C to BI), and GH equal to the one drawn

from B to AC ; it would be prov-

ed as before, that AP\AIv = BA.

AD + BC.CD, and I)n.AK =

AB.BC + AD.DC. Hence, AF.

AK : DH.AK, or (V. l) AF :

DII :: BA.AD-^BC.CD : AB.

BC + AD.DC. But the triani^les

AFC, DIIB are equiangular, liav-

in<x the rio;ht anojes Fand H, and

the angles ACF, DBH, each equal

(I. 16) to ALD; therefore (V. 3)

AF : AC : : DH : DB, and alternately AF : T3IT : : AC :

DB. Hence (IV. 7) the foregoing analogy becomes AC : BD

: : BA.AD + BC.CD : AB.BC-hAD.DC. Wherefore, the

diagonals, etc.

Scho. From this proposition and the last, when the sides of

a quadrilateral inscribed in a circle are given, we can find the

ratio of the diagonals and their rectangle, and thence (V, 15)

the diagonals themselves. Also, if the sides be given in num-

bers, we can compute the diagonals. Thus, let the sides taken

in succession round the figure be 50, 78, 104, and 120. Then,

the ratio of the diagonals will be that of 50 x 784-104 x 120 to

60x120-1-78x104; that is, 16880 to 14112, or 65 to 56, by-

dividing by 252. Again, the rectangle of the diagonals is 50 x

104-1-78x120, or 14560. But similar rectilineal figures are as

the squares of the corresponding sides, and consequently the

sides are as the square roots of the areas ; therefore, taking 65

and 56 as the sides of a rectangle, we have its area equal to

3640; and ^ 3640 is to 4/14560, or -^Z 3640 is to |/ (4x3640),

that is, 1 : 2 : : 65 : 130 : : 56

112 ; so that 130 and 112

are the diagonals.

Prop. VI. â€” Theor. â€” If in a straight line drawn through

the center of a circle^ and on the same side of (he center, (wo

povits be taken so that the radius is a mean proportional her-

tween their distances from, the cei^ter / two straight lines drawn

15

226 EXEECISES.

from, those points to any point whatever in the circumference^

are proportional to the segments into which the circumference

divides the straight line intercepted leticeen the same points.

Let ABC l)e a circle, and CAE a straight line drawn throngh

its center D ; if ED : DA : : DA : DF, and if BE, BF be

drawn from any point B of the circumference ; EB : BF : :

EA : AF.

Join AB, BD, Then, since DB is equal to DA, we have

(hyp.) ED : DB :: DB : DF.

The two triangles EDB, BDF,

therefore, have their sides about

the common angle D proportion-

al ; wherefore (V. 6) the angle

E is equal to FBD. Now the

angle liAD is equal (I. 20) to the

two angles E and EBA, and also

(I. 1, cor.) to ABD ; wherefore E and EBA are (I. ax. 1) equal

to ABD. From these take the equal angles E and FBD, and

(I. ax. 3) the remaining angles EBA, ABF are equal ; and

therefore (V. 4) in the triangle EBF, EB : BF : : EA : AF.

If, therefore, in a straight line, etc.

Cor. Join BC, and produce EB to G. Then, since ABC is

(III. 11) a right angle, it is equal to the two EBA, CBG.

From these equals take the equal angles ABF, EBA, and the

remainders FBC, CBG are equal ; and therefore (V. 4, 2d case)

EB : BF : : EC : CF. But it has been proved that EB :

BF : : EA : AF ; therefore (IV. 1) EA : AF : : EC : CF.

Hence, if the segments EA, AF be given, the point C may be

determined by the method shown in the third corollary to the

sixteenth proposition of the first book; and the circle ABC

may then be described, its diameter AC being determined.

Scho. The circle may also be determined in the following

manner: Since (hyp.) ED : DA : : DA : DF, by division,

EA : DA : : AF : DF ; whence, alternately and by division,

EAâ€” AF : AF : : AF : DF. Hence DF is a third proper-

tional to the difference of E A, AF, and to AF, the less ; and

thus the center D is determined. From the last analogy also

we obtained (IV. 11) EAâ€” AF : EA : : AF : AD; an anal-

ogy which serves the same purpose, since it shows that the ra-

EXERCISES. 227

dius of the circle is a fourth proportional to the difference of

the segments EA, AF, and to those segments themselves.

Prob. VII, â€” TnEOR. â€” The perpendiculars drawn from the

three angles of any triangle to the ojjposite sides, intersect one

another in the same point.

If the triangle be right-angled, it is plain that all the perpen-

diculars pass through the right angle. But if it be not right-

angled, let ABC be the triangle, and about it describe a circhi;

then, B and C being acute angles, draw ADE perpendicular to

BC, cutting BC in D, and the circumference in E; and make

DF equal to DE ; join BF and produce it, if necessary, to cut

AC, or AC produced in G ; BG is perpeixlicular to AC. Join

BE ; and because FD is equal to DE, the angles at D right

angles, and DB common to the two triangles FDB, EDB, the

angle FBD is equal (I. 3) to EBD ; but

(III. 18, cor. 1) CAD, EBD are also equal,

because they are in the same segment ;

therefore CAD is equal to FBD or GBC,

But the angle ACB is common to the tv o

triangles ACD, BCG; and therefore (I,

20, cor. 5) the remaining angles ADC,

BGC are equal ; but (const.) ADC is a

right angle; therefore also BGC is a right angle, and BG is

perpendicular to AC, In the same manner it would be shown

that a straight line CII, drawn through C and F, is perpendicu-

lar to AB. The three perpendiculars therefore all pass through

F ; wherefore, the perpendiculars, etc,

Scho. This limitation prevents the necessity of a different

ease, which would arise if the perpendicular AD fell without

the triangle. If the angle A be obtuse, the point F lies with-

out the circle, and BF, not produced, cutsAC produced. The

proof, however, is the same, and it is very easy and obvious.

Another easy and elegant proof, of which the following is an

outline, is given in Garnier's "Reciproques," etc., Theor. III.,

page 78 : Draw BG and CH perpendicular to AC and AB ;

join GH, and about the quadrilaterals AHFG and BHGC de-

8cribe circles, which can be done, as is easily shown ; draw

also AFD. Then the angles BAD, BCH are equal, each of

223 EXERCISKS.

them being equal (TIT, 18, cor. 1) to HGF ; and the angle ABO

"being coniinoii, ADB is equal (I. 20) to BHC, and is therefore

a right angle.

Prop. VTTT. â€” Theor. â€” From AB, the greater side of the tri-

angle ABC, cut off AD equal to AC, and join DC; draw AE

bisecting the vertical angle BAC, andjoiji DE ; d aw also AF

perpendicular to BC, and DG parallel to AE. Then (1.) the

angle DEB is eqtdvalent to the difference of the angles at the

base, ACB, ABC ; or of BAF, CAF, or of AEB, AEC ; and

DE is equal to EC ; (2.) the angles BCD, EAF are each equiv-

alent to half the same difference ; (3.) ADC or ACD is equiva-

lent to half the sum of the angles at the base, or to the comple-

ment of half the vertical angle ; (4.) BG is equivalent to the

difference of the segments BE, EC, made by the line bisecting

the vertical angle. ^

1. In the triangles AED, AEC, AD, AC are equal, AE com-

mon, and the contained angles equal ; therefore (I. 3) DE is

equal to EC, the angle ADE to ACE, and AED to AEC. But

(I. 20) because BD is produced, the

angle ADE is equivalent to B and

BED; therefore BED is the differ-

ence of B and ADE, or of B and

ACB. Also BED is the difference

of AEB, AED, or of AEB, AEC.

Again : ABF, BAF are equivalent

to ACF, CAF, each pair being (L

20, cor. 3) equivalent to a right angle. Take away ABF;

then, because the difference of ACF, ABF is BED, there re-

mains BAF equivalent to BED, CAF ; that is, BED is the

difference of BAI\ CAF.

Online Library → Lawrence S. (Lawrence Sluter) Benson → Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson → online text (page 18 of 21)