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Euclid.

Euclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools online

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Online LibraryEuclidEuclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools → online text (page 28 of 38)
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the same number of similar triangles.

Also these triangles shall have, each to each, the same ratio which
the polygons have to one another,

the antecedents being ABE, EBC, ECD, and the consequents
FGL, LGH,LHK:

and the polygon ABODE shall have to the polygon FGHKL the

duplicate ratio of that which the side AB has to the homologous

side FG. Because the triangle ABE is similar to the triangle FGL,

ABE has to FGL, the duplicate ratio of that which the side BE has

to the side GL : (vi. 19.)
for the same reason, the triangle BEC has to GLH the duplicate

ratio of that which BE has to GL :
therefore, as the triangle ABE is to the triangle FGL, so is the

triangle BEO to the triangle GLH. (v. 11.)
Again, because the triangle EBO is similar to the triangle LGH,
EBO has to LGH, the duplicate ratio of that which the side EC has

to the side LH:
for the same reason, the triangle ECD has to the triangle LHK, the
duplicate ratio of that which ^Chas to LH:
therefore, as the triangle EBO is to the triangle LGH, so is the tri-
angle ECD to the triangle LHK: (v. 11.)
but it has been proved,
that the triangle EBC is likewise to the triangle LGH, as the tri-
angle ABE to the triangle FGL :
therefore, as the triangle ABE to the triangle FGL, so is the triangle
EBC to the triangle L GH, and the triangle ECD to the triangle LHK t
and therefore, as one of the antecedents is to one of the consequents,

so are all the antecedents to all the consequents: (v. 12.)
that is, as the triangle ABE to the triangle FGL, so is the polvgon
AB ODE to the polygon FGHKL :
but the triangle ABE has to the triangle FGL, the duplicate ratio of

that which the side AB has to the homologous side FG; (vi. 19.)
therefore also the polygon ABODE has to the polygon FGHKL the
duplicate ratio of that which AB has to the homologous side FG.

Wherefore, similar polygons, &c. Q. E. D.
Cor. 1. In like manner it may be proved, that similar four-sided
figures, or of any number of sides, are one to another in the duplicate
ratio of their homologous sides : and it has already been proved in tri-
angles : (vi. 19.) therefore, universally, similar rectilineal figiu'es are to
one another in the duplicate ratio of their homologous sides.

Cor. 2. And if to AB, FG, two of the homologous sides, a third
proportional Mhe taken, (vi. 11.)



^80 Euclid's elements.

AJB has to M the duplicate ratio of that which AB has to FG:
(v. def. 10.)
but the four-sided figure or polygon upon AB, has to the four-
sided figure or polygon upon FG likewise the duplicate ratio of that
which AB has to FG: (VI. 20. Cor. 1.)

therefore, as AB is to M, so is the figure upon AB to the figure
uponi^G^: (v. 11.)

which was also proved in triangles : (vi. 19. Cor.)
therefore, universally, it is manifest, that if three straight lines be
proportionals, as the first is to the third, so is any rectilineal figure
upon the first, to a similar and similarly described rectilineal figure
upon the second.

PROPOSITION XXI. THEOREM.

Rectilineal figures ichich are similar to the same rectilineal Jigure, are
also similar to 07ie another.

Let each of the rectilineal figures A, B he similar to the rectilineal
figure C.

The figure A shall be similar to the figure B.



A Z>x



Because A is similar to (7,
they are equiangular, and also have their sides about the equal
angles proportional : (vi. def. 1.)

again, because B is similar to C,
they are equiangular, and have their sides about the equal angles

proportionals : (vi. def. 1.)
therefore the figures A, B are each of them equiangular to C, and
have the sides about the equal angles of each of them and of C pro-
portionals.

Wherefore the rectilineal figures A and B are equiangular,
(l. ax. 1.) and have their sides about the equal angles proportionals:
(v. 11.)

therefore A is similar to B, (vi. def. 1.)
Therefore, rectilineal figures, &c. q.e.d.

PROPOSITION XXII. THEOREM.

If four straight lines be proportionals, the similar rectilineal figures
similarly described upon them shall also be proportionals : and conversely,
if the similar rectilineal figures similarly described upon four straight lines
be proportionals, those straight lines shall be proportionals.

Let the four straight lines AB, CD, EF, GH be proportionals,

viz. A Bio CB, as EF io GH-,
and upon AB, CD let the similar rectilineal figures KAB, LCD be

similarly described ;
and upon EF, GH the Rimilar rectilineal figures MF^ NH, in like
manner :
the rectilineal figure KAB shall be to LCD, as MF io NH.



BOOK VI. PROP. XXIT, XXIII. 281

M

r V N S

-A \ \ rr\ -^ \Z\

ABCD EFGH Pli

To AB, CD take a third proportional X; (vi. 11.)

and to JEF, GH a third proportional O:

and because AB is to CI) as JEF to GIT,

therefore CD is to X, as G^^to O; (V. 11.)

•wherefore, ex sequali, as AB to X, so FF to : (V. 22.)

but as ^-B to X, so is the rectilineal figure KAB to the rectilineal

figure LCD,
and as EF to O, so is the rectilineal figure 3IF to the rectilineal
figure NH: (vi. 20. Cor. 2.)
therefore, as KAB to LCD, so is 3IF to NH. (v. 11.)
And if the rectilineal figure KAB be to LCD, as MF to iV:ffj
the straight line AB shall be to CD, as ii'i^ to G^jff.
Make as AB to CD, so ^i^to PR, (vi. 12.)
and upon BR describe the rectilineal figure SR similar and simi-
larly situated to either of the figures 3IF, NH: (vi. 18.)
then, because as AB to CD, so is FF to PR,
and that upon AB, CD are described the similar and similarly
situated rectilineals KAB, LCD,
and upon EF, PR, in like manner, the similar rectilineals MF, SR ;
therefore KAB is to LCD, as 3IF to SR :
but by the hypothesis KAB is to LCD, as 3IF to NH;
and therefore the rectilineal iLTF having the same ratio to each of the
two NH, SR,

these are equal to one another; (v. 9.)
they are also similar, and similiarly situated ;

therefore GH is equal to PR :

and because as ^5 to CD, so is EF to PR,

and that PR is equal to GH;

AB is to CD, as EF to GIL (v. 7.)

If therefore, four straight lines, &c. Q.E.D.

PROPOSITION XXIII. THEOREM.

Equiangular parallelograms have to one another the ratio which is
compounded of the ratios of their sides.

Let AC, CFhe equiangular parallelograms, having the angle J? CD
equal to the angle ECG.

Then the ratio of the parallelogram AC to the parallelogram CF,
shall be the same with the ratio which is compounded of the ratios of
their sides.

A D H




KLM



EUCLID S ELEMENTS.

Let BC, CG be placed in a straight line ;

therefore DC and CE are also in a straight line; (l. 14.)

and complete the parallelogram DG)

and taking any straight line K,

make as ^(7 to CG, so K to L ; (VL 12.)

and as Z)Cto CE, so make L to 31; (vi. 12.)

therefore, the ratios of K to L, and Z to iff, are the same with the

ratios of the sides,

viz. of i?Cto CG, and DC to CE-.
but the ratio of K to M is that which is said to be compounded of
the ratios of K to L, and L to M; (v. def. A.)
therefore K has to Jfthe ratio compounded of the ratios of the sides :
and because as i? C to CG, so is the parallelogram AC to the paral-
lelogram CH', (vi. 1.)

but as ^Cto CG', so is ^to i;
therefore K is to L, as the parallelogram AC to the parallelogram

CH: (V. 11.)
again, because as Z)C to CE, so is the parallelogram CH to the
parallelogram CF;

but as DC to CjE:, so is X to 3/;
wherefore L is to 31, as the parallelogram CH to the parallelogram
Ci^; (V. 11.)

therefore since it has been proved,
that as Kto L, so is the parallelogram ^Cto the parallelogram CH]
and as L to 3£, so is the parallelogram CH to the parallelogram CF;
ex ffiquali, ^is to 31, as the parallelogram AC to the parallelogram

CF: (V. 22.)
but JT has to iLTthe ratio which is compounded of the ratios of the

sides ;
therefore also the parallelogram ^Chas to the parallelogram CF,
the ratio which is compounded of the ratios of the sides.
Wherefore, equiangular parallelograms, &c. Q. E. D.

PROPOSITION XXIV. THEOREM.

Parallelograms about the diameter of any parallelogram, are similar to
the ichole, a7id to one another.

Let ABCD be a parallelogram, of which the diameter is AC;

and EG, HK parallelograms about the diameter.
The parallelograms EG, HK shall be similar both to the whole
parallelogram ABCD, and to one another.

A E B

/

/H




D K C

Because DC, GFare parallels,

the angle ^DCis equal to the angle AGF: (l. 29.)

for the same reason, because BC, EF are parallels,

the angle ABC is equal to the angle AEF:



BOOK VI. PROP. XXIV, XXV. 283

and each of the angles BCD, EFG is equal to the opposite angle
DAB, (I. 34.)

and therefore they are equal to one another :
wherefore the parallelograms ABCD, AEFG, are equiangular :

and because the angle ABCi& equal to the angle AEF,
and the angle 2?^ C common to the two triangles BAC, EAF,
they are equiangular to one another ;
therefore as AB to BC, so is AE to EF: (VI. 4.)
and because the opposite sides of parallelograms are equal to one
another, (I. 34.)

AB is to AD as AE to AG-, (v. 7.)

and DC to CB, as GF io FE ;
and also CD to DA, as EG to GA :
therefore the sides of the parallelograms ABCD, AEFG about the
equal angles are proportionals ;

and they are therefore similar to one another; (vi. def. 1.)
for the same reason, the parallelogram ABCD is similar to the

parallelogram FHCK:
wherefore each of the parallelograms GE, KHis similar to DB:
but rectilineal figures which are similar to the same rectilineal figure,
are also similar to one another: (VI. 21.)

therefore the parallelogram GE is similar to EH.
Wherefore, parallelograms, &c. Q.E.D.

PROPOSITION XXV. PROBLEM.

To describe a rectilineal figure which shall be similar to one, and equal
to another given rectilineal figure.

Let ABC he the given rectilineal figure, to which the figure to be
described is required to be similar, and D that to which it must be
equal.

It is required to describe a rectilineal figure similar to ABC, and
equal to D.

A

G H




M

Upon the straight line BC describe the parallelogram BE equal to
the figure ABC; (I. 45. Cor.)
also upon CE describe the parallelogram CM equal to D, (l. 45. Cor.)
and having the angle FCE equal to the angle CBL :
therefore ^Cand Ci^'are in a straight line, as also LE and EM:
(I. 21). and I. 14.)
between ^Cand Ci^ find a mean proportional GIT, (vi. 13.)
and upon Gil describe the rectilineal figure KG II similar and simi-
larly situated to the figure ABC. (VI. 18.)

Because i?C' is to G^J/as G^//to CF,



£84 Euclid's elements.

and that if three straight lines be proportionals, as the first is to
the third, so is the figure upon the first to the similar and similarly
described figure upon the second ; (vi. 20. Cor. 2.)
therefore, as ^Cto CF, so is the rectilineal figure ABC to KGH:
but as BC to CF, so is the parallelogram BF to the parallelogram

FF; (VT. 1.)

therefore as the rectilineal figure ABCk to KGH, so is the paral-
lelogram BF to the parallelogram FF: (v. 11.)
and the rectilineal figure ABCis equal to the parallelogram BF',

(constr.)
therefore the rectilineal figure KGH is equal to the parallelogram
FF: (V. 14.)

but FF is equal to the figure F ; (constr.)
■wherefore also KGHh equal to D: and it is similar to ABC.
Therefore the rectilineal figure KGH has been described similar to
the figure ^^C, and equal to D. q.e.f.

PKOPOSITION XXVI. THEOREM.

If two similar parallelog7-ams have a common a7igle, and be similarhj
situated; they are about the same diameter.

Let the parallelograms ABCD, AFFG be similar and similarly
situated, and have the angle DAB common.

ABCD and AFFG shall be about the same diameter.




For if not, let, if possible, the parallelogram BD have its diameter
AHC in a difi'erent straight line from AF, the diameter of the paral-
lelogram FG,

and let GF meet AHC in H;
and through ^draw -STZ" parallel to AD or BC;
therefore the parallelograms ABCD, AKHG being about the same
diameter, they are similar to one another ; (vi. 24.)

wherefore as DA to AB, so is GA to AK: (vi. def. 1.)

but because ABCD and AFFG are similar parallelograms, (hj^.)

as DA is to AB, so is GA to AF ;

therefore as GA to AF, so GA to AK; (v. 11.)

that is, GA has the same ratio to each of the straight lines AF, AK;

and consequently AK is equal to AF, (v. 9.)

the less equal to the greater, which is impossible :

therefore ABCD and AKHG are not about the same diameter :

wherefore ABCD and AFFG must be about the same diameter.

Therefore, if two similar, &c. Q. E. D.



BOOK VI. PROP. XXVII.



285



PROPOSITION XXVII. THEOREM.

Of all parallelograms applied to the same straight line, and deficient by
parallelograms^ similar and similarly situated to that which is described
upo?i the half of the line ; that which is applied to the half, and is similar
to its defect f is the greatest.

Let AB be a straight line divided into two equal parts in C;

and let the parallelogram AD be applied to the half A C, which is
therefore deficient from the parallelogram upon the whole line AB by
the parallelogram CU upon the other half CB :

of all the parallelograms applied to any other parts of AB, and
deficient by parallelograms that are similar and similarly situated to
CJS, AD shall be the greatest.

Let AF be any parallelogram applied to AX, any other part otAB
than the half, so as to be deficient from the parallelogram upon the
whole line AB by the parallelogram KII similar and similarly situ-
ated to CF :

DL E




C K B



AD shall be greater than AF.
First, let AK the base of AF, be greater than A C the half oiAB-.
and because CE is similar to the parallelogram UK, (hyp.)
they are about the same diameter : (vi. 26.)
draw their diameter DB, and complete the scheme :
then, because the parallelogram CF'is, equal to FE, (l. 43.)
add JT// to both:
therefore the whole CZTis equal to the whole KEi
but CH is equal to CG, (i. 36.)
because the base -4 C is equal to the base CB ;
therefore CG is equal to KE : (ax. 1.)
to each of these equals add CF;
then the whole AFh equal to the gnomon CHL : (ax. 2.)
therefore CE, or the parallelogram AD is greater than the paral-
lelogram AF.

Next, let AX the base of AFhe less than AC:



G F M H




thtn, the same construction being made, because BO is equal to CA,
therefore HM is equal to MG; (l. 31.]



286

therefore the parallelogram D-ETis equal to the parallelogram JjG]
(I. 36.)

wherefore DHh greater than LG :

but Dllk equal to DX; (l. 43.)

therefore DK is greater than LG:

to each of these add AL ;

then the whole AD is greater than the whole AF.

Therefore, of all parallelograms applied, &c. Q.B.D.

PROPOSITION XXVIII. PROBLEM.

To a given straight litie to ajjply a parallelogram equal to a given
rectilineal figure, and deficient by a parallelogram similar to a given paral-
lelogram : but the giveti rectilitieal figure to xchich the parallelogram to be
applied is to be equal, must not be greater than the parallelogram applied to
half of the given line, having its defeat similar to the defect of that which is
to be applied ; that is, to the given parallelogram.

Let AB be the given straight line, and Cthe given rectilineal figure,
to which the parallelogram to be applied is required to be equal, which
jfigure must not be greater (VI. 27.) than the parallelogram applied to
the half of the line, having its defect from that upon the whole line
similar to the defect of that which is to be applied ;

and let D be the parallelogram to which this defect is required to be
similar.

It is required to apply a parallelogram to the straight line AB,
which shall be equal to the figure C, and be deficient from the paral-
lelogram upon the whole line by a parallelogram similar to Z>.
Divide AB into two equal parts in the point JE, (I. 10.)
and upon JEB describe the parallelogram EBFG similar and simi-
larly situated to D, (vi. 18.)
and complete the parallelogram AG, which must either be equal to
C, or greater than it, by the determination.
1^ AGhe equal to C, then what was required is akeady done :
H G OF

L M




\^ a



E S i3 Iv N



for, upon the straight line ^5, the parallelogram ^ 6r is applied equal
to the figure C, and deficient by the parallelogram EF similar to D.
But, if -4 G^ be not equal to C, it is greater than it :
and FFh equal to -k6r; (I. 36.)
therefore FF also is greater than C.
Make the parallelogram KLMN equal to the excess of FF above
C, and similar and similarly situated to Dx (vi. 25.)

then, since D is similar to EF, (constr.)

therefore also KM i^ similar to EF, (vi. 21.)

let KL be the homologous side to EG, and X3/to GF:

and because EF is equal to C and KM together,



BOOK VI. PROP. XXVIII, XXIX. 2S1

JEF is greater than K3I;

therefore the straight line UG is greater than KL, and GF than Z3I:

make GX equal to ZK, and GO equal to L3f, (i. 3.)

and complete the parallelogram XGOP: (I. 31.)

therefore XO is equal and similar to KMi

but JST^/is similar to EF-,

wherefore also XO is similar to EF;

and therefore XO and EF are about the same diameter : (vi. 26.)

let GPB be their diameter and comxplete the scheme.

Then, because EF is equal to C and KM together,

and XO a part of the one is equal to KM o. part of the other,

the remainder, viz. the gnomon EliO, is equal to the remainder C:

(ax. 3.)
and because OH is equal to XS, by adding SIl to each, (I. 43.)

the whole 0J5 is equal to the whole XB :
but XB is equal to TE, because the base AE is equal to the base
EB; (1.36.)

wherefore also TJ^is equal to OB: (ax. 1.)
add XS to each, then the whole TS is equal to the whole, viz. to
the gnomon EliO :

but it has been proved that the gnomon EJRO is equal to C;

and therefore also TS is equal to C.

Wherefore the parallelogram TS, equal to the given rectilineal

figure C, is applied to the given straight line AB, deficient by the

parallelogram SJR, similar to the given one D, because SB is similar

to EF. (VI. 24.) Q.E.F.

PROPOSITION XXIX. PROBLEM.

To a given straight line to apply a parallelogram equal to a given recti-
lineal Jigure^ exceeding by a parallelogram similar to another given.

Let AB be the given straight line, and Cthe given rectilineal figure
to which the parallelogram to be applied is required to be equal, and D
the parallelogram to which the excess of the one to be applied above
that upon the given line is required to be similar.

It is required to apply a parallelogram to the given straight line
AB which shall be equal to the figure C, exceeding by a parallelogram
similar to D.



^5i



L M



\ \ Tx\

N P X



Divide AB into two equal parts in the point E, (l. 10.) and upon
EB describe the parallelogram EL similar and similarly situated to
jD: (VI. 18.)
and make the parallelogram GH equal to EL and C together, and
similar and similarly situated to D : (vi. 25.)

wherefore GMis similar to EL: (vi. 21.)



^88

let KJIhe the side homologous to FL, and KG to FU:

and because the paralleloo^ram GH is greater than FL,

therefore the side KH is greater than FLj

and KG than FF :

produce FL and FF, and make FL3I equal to -ff"^, and FFN to KG,

and complete the parallelogram 3fN:

3fN is therefore equal and similar to GS:

but GIT is similar to FL :

wherefore 3IN is similar to FL ;

and consequently FL and il/iV^are about the same diameter: (vi. 26.)

draw their diameter FX, and complete the scheme.

Therefore, since GH is equal to FL and C together,

and that GH is equal to ilifiV;

3fN is equal to FL and C :

take away the common part FL ;

then the remainder, viz. the gnomon NOL, is equal to C.

And because AE is equal to FB,

the parallelogram ^iV is equal to the parallelogram NB, (i. 36.)

that is, to B3I: (l. 43.)

add NO to each ;
therefore the whole, viz. the parallelogram AX, is equal to the
gnomon NOL :

but the gnomon NOL is equal to C;
therefore also AXi^ equal to C
Wherefore to the straight line AB there is applied the parallelo-
gram AX equal to the given rectilineal figure C, exceeding by the
parallelogram PO, which is similar to D, because PO is similar to
FL. (VI. 24.) Q.E.r.

PROPOSITION XXX. PROBLEM.
To cut a given straight line in extreme and mean ratio.

Let AB be the given straight line.
It is required to cut it in extreme and mean ratio.



lEB



Upon AB describe the square BC, (i. 46.)

and to ^C apply the parallelogram CD, equal to BC, exceeding by

the figure AD similar \o BC: (yi. 29.)

then, since BCh o. square,

therefore also AD is a square :

and because BCis equal to CD,

by taking the common part CF from each,

the remainder BF is equal to the remainder AD\

and these figures are equiangular.



BOOK VI. PROP. XXX, XXXI. 289

therefore their sides about the equal angles are reciprocally propor-
tional: (vi. 14.)

therefore, as FE to ED. &o AE to EB :

but FE is equal to AC, (l. 34) that is, to ^^; (def. 30.)

and ED is equal to AE-,

therefore as BA to AE, so k AE to EB :

but AB is greater than AE;
■wherefore AE is greater than EB : (v. 14.)
therefore the straight line ^i^ is cut in extreme and mean ratio in
E. (VI. def. 3.) Q.E.F.

Otherwise,

Let ^ J? be the given straight line.

It is required to cut it in extreme and mean ratio.



A C B

Divide AB in the point C, so that the rectangle contained by AB,
BC, may be equal to the square on A C. (ll. 11.)
Then, because the rectangle AB, BCis equal to the square on A C;
asBAto AC, so is AC to CB: (vi.l7.)
therefore AB is cut in extreme and mean ratio in C. (yi. def. 3.)

Q.E.F.

PROPOSITION XXXI. THEOREM.

In right-angled triafigles, the rectilineal figure described upon the side op-
posite to the right angle, is equal to the similar and similarly described figures
upon the sides contai?ii7ig the right angle.

Let ABC he a right-angled triangle, having the right angle BAC.
The rectilineal figure described upon 1? (7 shall be equal to the
similar and similarly described figures upon BA, A C.




Draw the perpendicular AD : (i. 12.)
therefore, because in the right-angled triangle ABC,
AD is drawn from the right angle at A perpendicular to the base BC,
the triangles ABD, ADC ^ve similar to the whole triangle ABC,
and to one another: (vi. 8.)

and because the triangle ABC is similar to ADB,
as CB to BA, so is BA to BD : (vi. 4.) ^
and because these three straight lines are proportionals,
as the first is to the third, so is the figure upon the first to the similar
and similarly described figure upon the second : (vi. 20. Cor. 2.)
therefore as CB to BD, so is the figure upon CB to the similar and

similarly described figure upon BA :
and inversely, as DB to BCy so is the figure upon BA to that upon
BC'. (V. B.)



290



for the same reason, as DC to CBy so is the figure upon CA to that

upon CB:
therefore as BD and DC together to BC, so are the figures upon
BA, AC io that upon B C: (v. 24.)

but BD and DC together are equal to BC',
therefore the figure described on BC is equal to the similar and
similarly described figures on BA, A C. (v. A.)
Wherefore, in right-angled triangles, &c. q.e.d.



PROPOSITION XXXII. THEOREM.

If two triangles which have two sides of the one proportional to two sides
of the othery be joined at one angle, so as to have their homologous sides
parallel to one another ; the remaining sides shall be in a straight line.

Let ABC, DCBJ be two triangles which have the two sides BA,
A C proportional to the two CD, DE,

viz. BA to AC, as CD to DE;
and let AB be parallel to DC, and ^C to DE.




Then J?C and CE shall be in a straight line.
Because AB is parallel to DC, and the straight line -4 C meets them,
the alternate angles BA C, A CD are equal ; (i. 29.)
for the same reason, the anirle CDE is equal to the angle A CD ;
wherefore also BACis equal to CDE: (ax. 1.)
and because the triangles ABC, DCEh3.\e one angle at A equal to
one at D, and the sides about these angles proportionals,
viz. BA to A C, as CD to DE,
the triangle ABC is equiangular to DCE: (vr. 6.)
therefore the angle ABC is equal to the angle DCE:
and the angle BA C was proved to be equal to A CD ;
therefore the whole angle ACE is equal to the two angles ABC,
BAC: (ax. 2.)
add to each of these equals the common angle A CB,
then the angles A CE, A CB are equal to the angles ABC, BA C, A CB :
but ABC, BAC, ACB are equal to two right angles: (l. 32,)
therefore also the angles A CE, A CB are equal to two right angles :
and since at the point C, in the straight line A C, the two straight
lines BC, CE, which are on the opposite sides of it, make the adjacent
angles A CE, A CB equal to two right angles ;

therefore ^Cand CE are in a straight line. (l. 14.)



Online LibraryEuclidEuclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools → online text (page 28 of 38)