Euclid.

# A text-book of Euclid's Elements : for the use of schools : Books I-VI and XI online

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Online LibraryEuclidA text-book of Euclid's Elements : for the use of schools : Books I-VI and XI → online text (page 25 of 27)
Font size liave four faces ; or, if two faces are parallel, it must at least have
Jive faces.

14. A prism is a solid figure bounded by plane faces,
of which two that are opposite are similar and equal
polygons in parallel planes, and the other faces are paralle-
lograms.

The polygons are called the ends of the prism. A prism is
said to be right if the edges formed by each pair of adjacent parallel-
ograms are perpendicular to the two ends; if otherwise the prism is
oblique.

IT). A parallelepiped is a solid figure bounded by
three pairs of parallel plane faces.

Fig. 1. Fig. 2.

A parallelepiped may be rectangular as in fig. 1, or oblique as in
fig. 2.

DEFINITIONS.

389

16. A pyramid is a solid figure bounded by plane
faces, of which one is a polygon, and the rest are triangles
liaving as bases the sides of the polygon, and as a com
mon vertex some point not in the plane of the polyf^on.

The polygon is called the base of the pyramid.

A pyramid having for its base a regular polygon is said to be
right when the vertex lies in the straight line drawn perpendicular
to the base from its central point (the centre of its inscribed or cir-
cumscribed circle).

17. A tetrahedron is a pyramid
on a triangular base : it is thus con-
tained hj four triangular faces.

18. Polyhedra are classified according to the number
of their yaces .â€˘

thus a hexahedron has six faces ;
an octahedron has eight faces ;
a dodecahedron has hcelve faces.

19. Similar polyhedra are such as have all their solid
angles equal, each to each, and are bounded by the same
number of similar faces.

20. A Polyhedron is regular when its faces ai-e similar
and equal r(jgular polygons.

;i9o

KUCLID S KI>KMKN"I'S.

21. It will be proved (see page 425) that there can
only he. Jive regular polyhedra.

They are defined as follows.

(i) A regular tetrahedron is a

solid figure bounded by four pi;
faces, which are equal and e((ui
lateral triangles.

(ii) A cube is a solid figure
bounded by six plane faces, which
are equal squares.

(iii) A regular octahedron is a
solid figure bounded by eight plane
faces, which are equal and equilateral

38.

(iv) A regular dodecahedron is /
a solid figure bounded by tvjelve plane
faces, which are equal and regular
pentagons.

DEFINITIONS.

391

(v) A regular icosahedron is
a solid figure bounded by twenty
plane faces, which ;ire equal and
equilateral fricnff/fcs.

Solids of Revolution.

22, A sphere is a solid figure described by the revo-
lution of a semicircle about its diameter, which remains
fixed.

The axis of the sphere is the fixed straight line about which the
semicircle revolves.

The centre of the sphere is the same as the centre of the semi-
circle.

A diameter of a sphere is any straight line which passes through
the centre, and is terminated both ways by the surface of the
sphere.

23. A right cylinder is a solid
figure described by the revolution of
a rectangle about one of its sides
which remains fixed.

The axis of the cylinder is the fixed straight line about which the
rectangle revolves.

The bases, or ends of the cylinder are the circular faces described
by the two revolving opposite sides of the rectangle.

KDCLID'S ELKMKN'rs

24. A right cone is a solid figure
described by the re\'olution of a right-
angled triangle about one of the sides
containing the right angle which re-
mains fixed.

The axis of the cone is the fixed straight line about which the
triangle revolves.

The base of the cone is the circular face described by that side
which revolves.

The hypotenuse of the right-angled triangle in any one of its
positions is called a generating line of the cone.

25. Similar cones and cylinders are those which have
their axes and the diameters of their bases proportionals.

BOOK XI. PROP. 1. 393

Proposition 1. Theorem.

One part of a straight line cannot he in a plane and an-
other part outside it.

If possible, let AB, part of the st. line ABC, be in the
plane PQ, and the part BC without it.

Then since the st. line A B is in the plane PQ,

.". it can be produced in that plane, i. Post. 2.

Produce AB to D;
and let any other plane which passes through AD be turned

Then because the points B and C are in this plane,

.'. the st. line BC is in it: i. Def. 5.

.â€˘. ABC and ABD are in the same plane and are both

St. lines ; which is impossible. i. Def. 3.

.'. the st. line ABC has not one part AB in the plane PQ,

and another part BC outside it. q. e. d.

Note. This proposition scarcely needs proof, for the truth of it
follows almost immediately from the definitions of a straight line
and a plane.

It should be observed that the method of proof used in this and
the next proposition rests upon the following axiom.

If a plane of unlimited extent turns about a fixed straifjht line as
an axis, it can be made to 2)ass through any point in space.

394 Euclid's et-kmentr.

Proposition 2. Theorem,

Any two straight lines which cut one another are in one
plane: and any three straight lines, of which each pair inter-
sect one another, are in one plnnie.

Let the two st. lines AB and CD intersect at E;
and let the st. line BC be drawn cutting AB and CD at B
and C:

then (i) AB and CD shall lie in one plane,
(ii) AB, BC, CD shall lie in one plane,
(i) ' Let any plane pass through AB ;

and let this plane be turned about AB until it passes
through C.

Then, since C and E are points in this plane,
.'. the whole st. line CED is in it. I. Def. 5 and xi. L
That is, AB and CD lie in one plane.

(ii) And since B and C are points in the plane which
contains AB and CD,

.'. also the st. line BC lies in this plane. Q. e. d.

Corollary. One, and only one, plane can he made to
pass through tivo given intersecting straight lines.

Hence the position of a plane is fixed,

(i) if it passes through a given straight line and a given point
outside it; ^ a:, p. 393.

(ii) if it passes through two intersecting straight lines ; xi. 2.

(iii) if it passes through three points not collinear ; xi. 2.

(iv) if it passes through two parallel straight lines. i. Def. 25,

BOOK XI. PROP. 3.

395

Proposition 3. Theorem.

If two 'planefi cut one another their common section is a
straight line.

C

Let the two planes XA, CY cut one another, and let BD be
their common section :

then shall BD be a st. line.
For if not, from B to D in the plane XA draw the st. line
BED;

and in the plane CY draw the st. line BFD.

Tlien the st. lines BED, BFD have the same extremities;

.'. they include a space;

but this is impossible.

.'. the common section BD cannot be otherwise than a st.

line. Q. E. D.

Or, more briefly thus â€”
Let the planes XA, CY cut one another, and let B and D be
two points in their common section.
Then because B and D are two points in the plane XA,
.'. the st. line joining B, D lies in that plane, i. Def. 5.
And because B and D are two points in the plane CY,
.*. the st. line joining B, D lies in that plane.
Hence the st. line BD lies in both planes,
and is therefore their common section.
That is, the common section of the two planes is a straight
line. Q. E. D.

39U kuclid's elements.

Proposition 4. Theorem. [Alternative Proof.]

If a straight line is perpendicular to each of two straight
lines at their point of intersection, it shall also be perpen-
dicular to the plane in which they lie.

Let the straight line AD be perp. to each of the st.
lines AB, AC at A their point of intersection:
then shall AD be perp. to the plane in which AB and
AC lie.

Produce DA to F, making AF equal to DA.

Draw any st. line BC in the plane of AB, AC, to cut
AB, AC at B and C;

and in the same plane draw through A any st. line AE to cut
BC at E.

It is required to prove that AD is perp. to AE.
Join DB, DE, DC ; and FB, FE, FC.

Then in the AÂ« BAD, BAF,
because DA = FA,
and the common side AB is perp. to DA, FA
.-. BD= BF.
Similarly CD = CF.
Now if the ABFC be turned about its base BC until
the vertex F comes into the plane of the A BDC,

then F will coincide with D,

since the conterminous sides of tlie triangles are equal, i. 7.

.*. EF will coincide with ED,

that is, EF^ED.

Constr

I. 4.

BOOK XI. PROP. 4.

397

Hence in the a ^ DAE, FAE,
since DA, AE, ED = FA, AE, EF respectively,

.-. the L DAE = the l FAE. i. 8.

Tliat is, DA is perp. to AE.
Similarly it may be shewn that DA is perp, to every
St. line which meets it in the plane of AB, AC ;

.'. DA is perp. to this plane. q.e.d.

Pro POSITION -i. Theorem. [Euclid's Proof.]

If a straight line is perpendicular to each of tivo straight
lines at their point of intersection^ it shall also he 'perpen-
dicular to the plane in which they lie.

Let the st. line EF be perp. to each of the st. lines
AB, DC at E their point of intersection :
then shall EF be also perp. to the plane XY, in which
AB and DC lie.

Make EA, EC, EB, ED all equal, and join AD, BC.
Through E in the plane XY draw any st. line cutting
AD and BC in G and H.

Take any pt. F in EF, and join FA, FG, FD, FB, FH, FC.

Then in the a ^ AED, BEC,

because AE, ED = BE, EC respectively, Co7tstr.

and the L AED = the L BEC ; i. 15.

.*. AD = BC, and the :_ DAE = the L CBE. i. 4.

39Â«

EUCLID'S ELEMENTS.

F

In the A Â« AEG, BEH,

because the l GAE = the l HBE,

Proved.

and the L AEG - the l BEH,

I. 15.

and EA- EB;

Constr.

.-. EG = EH, and AG = BH.

I. 26.

Again in the a ^ FEA, FEB,

because EA = EB,

and the common side FE is perp. to EA, EB

/%;.

.-. FA-FB.

r. 4.

Similarly FC-FD.

Again in the A** DAF, CBF,
because DA, AF, FD ^ CB, BF, FC, respectively,

.â€˘. the ÂŁ. DAF^the^CBF. l 8.

And in the a ^ FAG, FBH,
because FA, AG, = FB, BH, respectively,

and the ^ FAG = the /.FBH, Proved.

.-. FG = FH. I. 4.

Lastly in the a Â» FEG, FEH,
because FE, EG, GF = FE, EH, H F, respectively,
.-. the /.FEG -the l FEH;
that is, FE is perp, to GH.

r. 8.

Similarly it may be shewn that FE is perp. to evei-y
St. line which meets it in the plane XY,

.*. FE is perp. to this plane. xi. Be/.

Q.E.D.

BOOK XI. PROP. 5.

399

Proposition 5. Theorem.

If a straight line is i^erfendicular to each of three con-
current straight lines at their point of intersection, these
three straight lines shall he in one plane.

Let the straight line AB be perpendicular to each of
the straight lines BC, BD, BE, at B their point of inter-
section :

then shall BC, BD, BE be in one plane.

Let XY be the plane which passes through BE, BD ; xi. 2.

and, if possible, suppose that BC is not in this plane.

Let AF be the plane which passes through AB, BC ;
and let the common section of the two planes XY, AF be the
St. line BF. xi. 3.

Then since AB is perp. to BE and BD,
and since BF is in the same plane as BE, BD,

.*. AB is also perp. to BF. xi. 4.

But AB is perp. to BC; Hyp-

.'. the L^ ABF, ABC, which are in the same plane, are
both rt. angles ; which is impossible.

.'. BC is not outside the plane of BD, BE :
that is, BC, BD, BE are in one plane.

Q.E.D.

4U)

EL'CLIDS ELEMENTS.

Proposition 6. Theorem.

If two straight lines are 'perpeMdicular to the same j^lane^
they shall be jmralhl to one another.

X

Let the st. lines AB, CD be perp. to the plane XY :

then shall AB and CD be par'.*

Let AB and CD meet the plane XY at B and D.

Join BD;

and in the plane XY draw DE perp. to BD, making DE

equal to AB.

Then since AB is perp, to the plane XY, ^%i^-

.*. AB is also perp. to BD and BE, which meet it in that
plane ; xi. Be/. 1.

that is, the l ^ ABD, ABE are rt. angles.
Similarly the z. ** CDB, CDE are rt. angles.
Now in the A '^ ABD, EDB,
becauso AB, BD =^ ED, DB, respectively, Constr.
and the /. ABD = the z. EDB, being rt. angles ;

.'. AD = EB. I. 4.

Again in the a ^ ABE, EDA,
because AB, BE - ED, DA, respectively,
and AE is common ;
.'. the z. ABE -the ^ EDA. i. 8.

* Note. In order to shew that AB*and CD are parallel, it is
necessary to prove that (i) they are in the same plane, (ii) the angles
ABD, CDB, are supplementary.

BOOK XT. PROP. 7. 401

But the z_ ABE is a rt. angle ; Proved.

.'. the z_ EDA is a rt. angle.
But the L EDB is a rt. angle by construction,

and the i. EDO is a rt. angle, since CD is perp. to the

plane XY. Ji^yp-
Hence ED is perp. to the three lines DA, DB, and DC;

.'. DA, DB, DC are in one plane. xi. 5.

But A B is in the plane which contains DA, DB ; xi. 2.

.'. AB, BD, DC are in one plane.

And each of the z_ ^ ABD, CDB is a rt. angle ; Hyp.

.'. AB and CD are par'. i. 28.

\ Q.E.D.

Proposition 7. Theorem.

If two straight lines are parallel, the straight line which
joiris any point hi one to any point in the other is in the
same plane as the 2^ci'f'cillels.

Let AB and CD be two par' st. lines,
and let E, F be any two points, one in each st. line :
then shall the st. line which joins E, F be in the same
plane as AB, CD.

For since AB and CD are par',

.'. they are in one plane. i. De/. 25.

And since the points E and F are in this plane,
.*. the St. line which joins them lies wholly in this plane.

I. Def. 5.
That is, EF is in the plane of the par'^ AB, CD.

Q.E.D,
H. K. 26

402

euclid'8 elements.

Proposition 8. Theorem.

If two straight lines are parallel, and if one of them
is perpendicular to a plane, then the other shall also be per-
pendicular to the same plane.

Let AB, CD be two par' st. lines, of which AB is perp.
to the plane XY :

then CD shall also be perp. to the same plane.
Let AB and CD meet the plane XY at the points B, D.
Join BD ;
and in the plane XY draw DE perp. to BD, making DE equal
to AB.

Then because AB is perp. to the plane XY, Hyp-

.'. AB is also perp. to BD. and BE, which meet it in that
plane ; xi. JJef 1.

that is, the z. " ABD, ABE are rt. angles.

Now in the A** ABD, EDB,
because AB, BD = ED, DB, respectively, Constr.

and the z. ABD â€” the z. EDB, being rt. angles;

Again in the A^ ABE, EDA,
because AB, BE - ED, DA respectively,
and AE is common ;
.-. the L ABE = the i. EDA. i. 8.

BOOK XI. PROP. 8. 403

But the z_ ABE is a rt. angle ; Proved.

.'. the L EDA is a rt. angle :
that is, ED is perp. to DA.
But ED is also perp, to DB : Constr.

.*. ED is perp. to the plane containing DB, DA. xr. 4
And DC is in this plane ;
for both DB and DA are in the plane of the par'*' AB, CD.

XI. 7 ..
:. ED is also perp. to DC ; xi. Bef. 1.

that is, the z_ CDE is a rt. angle.
Again since AB and CD are par', J^yp-

and since the z. ABD is a rt. angle,
.". the z. CDB is also a rt. angle. i. 29.

.'. CD is perp. both to DB and DE ;
.â– . CD is also perp. to the plane XY, Avliich contains
DB, DE. XI. 4.

Q.E.D.

EXERCISES.

1. The perpendicular is the least straight line that can be drawn
from an external point to a plane.

2. Equal straight lines drawn from an external point to a plane
are equally inclined to the perpendicular drawn from that point to
the plane.

3. Shew that two observations with a spirit-level are sufficient to
determine if a plane is horizontal : and prove that for this purpose
the two positions of the level must not be parallel.

4. What is the locus of points in space which are equidistant
from two fixed points ?

5. Shew how to determine in a given straight line the point
which is equidistant from two fixed points. When is this im-
possible ?

6. If a straight line is parallel to a plane, shew that any plane
passing through the given straight line will have with the given plane
a common section which is parallel to the given straight line.

26-2

404

KITCLID'S ELKMKNTS.

Proposition 9. Theorkm.

I'wo straight lines which are 2)(^rallel to a third straight
line are parallel to one another.

Let the st. lines AB, CD be each par' to the st. line PQ :
then shall AB be par' to CD,

T. If AB, CD and PQ are in one plane, tlio proposition has

II. But if AB, CD and PQ are not in one plane,

in PQ take any point G ;
and from G, in the plane of the par'" AB, PQ, draw GH

perp. to PQ; i. 11.

also from G, in the plane of the par'" CD, PQ, draw

GK perp. to PQ. i. 11.

Then because PQ is perp. to GH and GK, Constr.

.'. PQ is perp. to the plane HGK, which contains them.

XI. 4.
But AB is par' to PQ ; ^^W-

.'. AB is also perp. to the plane HGK. xi. 8.

Similarly, CD is perp. to the plane HGK.

Hence AB and CD, being perp. to the same plane, are par*
to one another, xi. 6.

Q.K.D.

BOOK XL PROP. 10.

405

Proposition 10. Theorem.

//' two intersecting straight lines are respectively parallel
to two other intersectifig straight lines' not in the same plane
with them, then the first pair and the second pair shall con-
tain equal angles. /

Let the st. lines AB, BC be respectively par' to the st.
lines DE, EF, which are not in the same plane with them :
then shall the z. ABC = the i. DEF.

In BA and ED, make BA equal to ED ;

and'in BC and EF, make BC equal to EF.
Join AD, BE, CF, AC, DF.

Then because BA is equal and par' to ED,

Hyp. and Constr.
.'. AD is equal and par' to BE. i, 33.

And because BC is equal and par' to EF,

.'. CF is equal and par' to BE. i. 33.

,'. AD is equal and par' to CF ; xi. 9.

hence it follows that AC is equal and par' to DF. i. 33.
Then in the A^ ABC, DEF,
because AB, BC, AC - DE, EF, DF, respectively,

.â€˘. tlie L ABC = the _ DEF. i. 8.

Q.E.D.

406

EUCLID S ELEMENTS.

Proposition 11. Problem.

2^0 draiv a straight line 2>^tyendicular to a given flane
from a given point outside it.

Let A be the given point outside the plane XY.
It is required to draw from A a st. line perp
plane XY.

Draw any st. line BC in the plane XY ;
and from A draw AD perp. to BC
Then if AD is also perp.
required is done.

But if not, from D draw
to BC;

and from A draw AF perp. to DE.
Then AF shall be perp. to the plane XY.

Through F draw FH par' to BC.
Now because CD is perp. to DA and DE,
.'. CD is perp. to the plane containing DA, DE
And HF is par' to CD ;
.'. HF is also perp. to the plane containing DA,
And since FA meets HF in this plane
.". the /. HFA is a rt. angle ;
that is, AF is perp. to FH.
And AF is also perp. to DE;
.'. AF is perp. to the plane containing FH
that is, AF is perp. to the plane XY.

to the

I. 12.
to the plane XY, what was

DE in the plane XY perp.
L 11.
I. 12.

I. 31.

Constr,

XI. 4.

DE, XI. 8.
XI. Def. 1.

Co7istr.
DE;

Q.E.F.

book xi. prop. 12.
Proposition 12. Problem.

407

To draw a straight line perpendicular to a given plane
from a given point in the plane.

D B

Let A be the given point in the plane XY.
It is required to draw from A a st. line perp. to the
plane XY.

From any point B outside the plane XY draw BC perp.
to the plane. xi. 11.

Then if BC passes through A, what was required is
done.

But if not, from A draw AD par' to BC, i. 31.

Then AD shall be the perpendicular required.

For since BC is perp. to the plane XY, Constr.

and since AD is par' to BC, Constr.

.'. AD is also perp. to the plane XY. xi. 8.

Q.E.F.

EXERCISES.

1. â–  Equal straight lines drawn to meet a plane from a point
without it are equally inclined to the plane.

2. Find the locus of the foot of the perpendicular drawn from a
given point upon any plane which passes through a given straight
line.

3. From a given point A a perpendicular AF is drawn to a plane
XY; and from F, FD is drawn perpendicular to BC, any line in
that plane: shew that AD is also perpendicular to BC.

408

KUCLID'S ELEMENTS.

Proposition 13. Theorem.

Only one perpendicular can he drawn to a given plane
from a given point either in the plane or outside it. '

F

B C

\ â– !

Y

y^

\ 1

/

/

V

/p

A /

Case I. Let the given point A be in the given plane XY ;
and, if possible, let two perps. AB, AC be drawn from A to
the plane XY.

Let DF be the plane which contains AB and AC ; and

let the st. line DE be the common section of the planes DF

and XY. XI. 3.

Then the st. lines AB, AC, AE are in one plane.

And because BA is perp. to the plane XY, Ji^yp-

.'. BA is also perp. to AE, which meets it in this plane ;

XL Def. L
that is, the ^ BAE is a rt. angle.
Similarly, the ^ CAE is a rt. angle.
.*. the I. ^ BAE, CAE, which are in the same plane, are equal
to one another.

Which is impossible.
.'. two perpendiculars cannot be drawn to the plane
XY from the point A in that plane.

Case I J. Let the given point A be outside the plane XY.
Then two perp^ cannot be drawn from A to the plane ;
for if there could be two, they would be par', xi. 6.
which is absurd. q.e.d.

BOOK XI. PROP. 14.

409

Proposition 14. Theorem.

Planes to which the same straight line is perpendicular
are parallel to one another.

! /

v

\

1

[

D

B

Let the st. line AB be perp. to each of the planes CD, EF :

then shall these planes be par'.

For if not, they will meet when produced.

If possible, let the two planes meet, and let the st.

line GH be their common section. xi. 3.

In G H take any point K ;

and join AK, BK.

Then because AB is perp. to the plane EF,
.'. AB is also perp. to BK, which meets it in this plane;

XI. Be/. 1.
that is, the z_ ABK is a rt. angle.
Similarly, the z. BAK is a rt. angle.
.'. in the A KAB, the two L ^ ABK, BAK are together equal to
two rt. angles ;

which is impossible. I. IT.

.'. the planes CD, EF, though produced, do not meet :

that is, they are par'. q.e.d.

410

eucud'8 elements.

Proposition 15. Theorem.

If two intersecting straight lines are parallel respectively
to two other intersecting straight lines which are not in
tlie same plane with them^ then the plane containing the
first pair shall be parallel to the plane containing the second
jiair.

Let the st. lines AB, BC be respectively par' to the
st. lines DE, EF, which are not in the same plane as
AB, BC :

then shall the plane containing AB, BC be par' to the
l^lane containing DE, EF.

From B draw BG perp. to the plane of DE, EF ; xi. 11.
and let it meet that plane at G,
Through G draw GH, GK par' respectively to DE, EF. i. 31.

Then because BG is perp. to the plane of DE, EF,
,'. BG is also perp. to GH and GK, which meet it in that
plane : xi. Def. 1.

that is, eacli of .the l"" BGH, BGK is a rt. angle.

Now because BA is par' to ED, I^VP-

and because GH is also par' to ED, Constr.

.'. BA is par' to GH. xi. 9.

And since the /.BGH is a rt. angle ; Proved.

:. the L ABG is a rt. angle. I. 29.

Similarly the z. CBG is a rt. angle.

BOOK XI. PROP. 16.

411

Then since BG is perp. to each of the st. lines BA, BC,
.'. BG is perp. to the plane containing them. xi. 4.
But BG is also perp, to the plane of ED, EF : Gonstr.
that is, BG is perp. to the two planes AC, DF :

.*. these planes are par'. xi. 14.

Q.E.D.

Proposition 16. Theorem.

//* two parallel planes are cut hy a third plane their
common sections 'with it shall be parallel.

N

'F

c

M

i^

\

\

\

D

b.

r

^

G

Let the par' planes AB, CD be cut by the plane EFHG,
and let the st. lines EF, GH be their common sections
with it :

then shall EF, GH be par'.

For if not, EF and GH will meet if produced.

If possible, let them meet at K.

Then since the whole st. line EFK is in the plane AB, xi. 1.

and K is a point in that line,

.'. the point K is in the plane AB.

Similarly the point K is in the plane CD.

Hence the planes AB, CD when produced meet at K ;

which is impossible, since they are par'. Hyp.

.'. the st. lines EF and GH do not meet;
and they are in the same plane EFHG ;

.*. they are par'. i. De/. 25.

Q.E.D.

412 euclid'8 elements.

Proposition 17. Theorem.

Straight lines which are cut by parallel planes are cut
proportionally.

H

Let the st. lines A B, CD be cut by the three par' planes
GH, KL, MN at the points A, E, B, and C, F, D :
then shall AE : EB :: CF : FD.

and let AD meet the plane KL at the point X :

join EX, XF.

Then because the two par' planes KL, MN are cut by
the plane ABD,

.". the common sections EX, BD are par'. xi. 16.

and because the two par' planes GH, KL are cut by tlie
plane DAC,

.*. the common sections XF, AC are par'. xi. IG.

And because EX is par' to BD, a side of the A ABD,

.'. AE : EB :: AX : XD. vi. 2.

Again because X F is par' to AC, a side of the A DAC,

Online LibraryEuclidA text-book of Euclid's Elements : for the use of schools : Books I-VI and XI → online text (page 25 of 27)