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

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.-. AX : XD :: CF : FD. vi. 2.

Hence AE : EB :: CF : FD. v. 1.

C,).E.D.

Definition. One plane is perpendicular to another
plane, when any straight line drawn in one of the planes
perpendicular to their common section is also perpendicular
to the otlier plane. [Book xi. Def. C]



BOOK XI. PROP. 18.



41.3



Proposition 18. Theorem.

If a straight line is 2yerpendicidar to a plane, then every
plane which passes through the straight line is also
dicular to the given plane.




Let the st. line AB be perp. to the plane XY ;
and let DE be any plane passing through AB :
then shall the plane DE be perp. to the plane XY.
Let the st. line CE be the common section of the planes
XY, DE. XI. 3.

From F, any point in CE, draw FG in the plane DE
perp. to CE. i. 11.

Then because AB is perp. to the plane XY, Hyp.
:. AB is also perp. to CE, which meets it in that plane,

XI. Def. 1.

that is, the ^ ABF is a rt. angle.

But the z. GFB is also a rt. angle ; Constr.

:. GF is par' to AB. I. 28.

And AB is perp. to the plane XY, Hyp.

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

Hence it has been shewn that any st. line G F drawn in

the plane DE perp. to the common section CE is also perp.

to the plane XY.

.'. the plane DE is perp. to the plane XY. xi. Def. 6.

Q.E.D.



EXERCISE.



Sheio that tivo planes are perpendicular to one another tohen the
dihedral angle formed by them is a right angle.



414



ECCLIDS ELEMENTS.



Proposition 19. Theorem.

If two intersecting iilanes are each perpendicular to a
third plane, their common section shall also be perpendicular
to that plane.




Let each of the planes AB, BC be perp. to the plane
ADC, and let BD be their common section :

then shall BD be perp. to the plane ADC.
For if not, from D draw in the plane AB the st. line DE
perp. to AD, the common section of the planes ADB, ADC :

I. 11.
and from D draw in the plane BC the st. line DF perp.
to DC, the common section of the planes BDC, ADC.
Then liecause the plane BA is perp. to the plane ADC,

Hyp.

and DE is drawn in the plane BA perp. to AD the common

section of these planes, Constr.

:. DE is perp. to the plane ADC. xi. Dp/. 6.

Similarly DF is perp. to the plane ADC.

.'. from the point D two st. lines are drawn perp. to the

plane ADC ; which is impossible. xi. 13.

Hence DB cannot be otherwise than perp. to the plane ADC.

Q.E.D.



BOOK XI. PROP. 20. 415



Proposition 20. Theorem.



Of the three plane angles which form a trihedral angle,
any two are together greater than the third.




Let the trihedral angle at A he fonnecl by the three
plane L^ BAD, DAC, BAG :

then shall any two of them, such as the l ^ BAD, DAC, he
together greater than the third, the l BAG.

Case I. If the l. BAG is less than, or equal to, either
of the /^ BAD, DAG;

it is evident that the /. ^ BAD, DAG are together greater
than the l BAG.

Case II. But if the l BAG is greater than either of the
L^ BAD, DAG;

then at the point A in the plane BAG make the z. BAE equal
to the L BAD ;

and cut off AE equal to AD.
Through E, and in the plane BAG, draw the st. line BEG
cutting AB, AG at B and G :

join DB, DG.

I Then in the a « BAD, BAE,

since BA, AD = BA, AE, respectively, Constr.

and the /.BAD = the L BAE ; Constr.

.'. BD = BE. I. 4.

Again in the ABDG, since BD, DG are together greater
than BG, I. 20.

andBD^BE, Proved.

.'. DG is greater tlian EC



4\i>



EUCLID S ELEMENTS.
D




And in the a ^ DAC, EAC,
because DA, AC = EA, AC respectively,
Imt DC is greater than EC ;



the L. DAC is greater than the



EAC.



Cjonstr.

Proved.

I. 25.



But the /.BAD =the z. BAE ; Constr.

:. tlie two _ ** BAD, DAC are together greater than the

BAC. Q.E.D.



Proposition 21. Theorem.

Every [convex) solid angle is formed hy plane angles
which are together less than four right angles.




Let the solid angle at S be formed by the plane L ^ ASB,
BSC, CSD, DSE, ESA:
then shall the sum of these plane angles be less than four



rt. angles.



BOOK XI. PROP. 21. 417

For let a plane XY intersect all the arms of the plane
angles on the same side of the vertex at the points A, B, C,
D, E : and let AB, BC, CD, DE, EA be the common sections
of the plane XY with the planes of the several angles.
Within the polygon ABODE take any point O ;
and join O to each of the vertices of the polygon.
Then since the a. ^ SAE, SAB, EAB form the trihedral
angle A,

.*. the L ^ SAE, SAB are together greater than the l EAB ;

XI. 20.
that is,
the L ^ SAE, SAB are together greater than the l ^ OAE, OAB.

Similarly,

the L. ^ SB A, SBC are together greater than the l ^ OB A, OBO:
and so on, for each of the angular points of the polygon.
Thus by addition,
the sum of the base angles of the triangles whose vertices
are at S, is greater than the sum of the base angles of
the triangles whose vertices are at O.
But these two systems of triangles are equal in number ;
/. the sum of all the angles of the one system is equal to the
sum of all the angles of the other. .
It follows that the sum of the vertical angles at S is
less than the sum of the vertical angles at O.

But the sum of the angles art O is four rt. angles ;
/.the sum of the angles at S is less than four rt. angles.

Q.E.D.

'Note. This proposition was not given in this form by Euclid, who
established its truth only in the case of trihedral angles. The above
demonstration, however, applies to all cases in which the polygon
ABODE is convex, but it must be observed that without this condition
the proposition is not necessarily true.

A solid angle is convex when it lies entirely on one side of each
of the infinite planes which pass through its plane angles. If this is
the case, the polygon ABODE will have no re-entrant angle. And it
is clear that it would not be possible to apply xi. 20 to a vertex at
which a re-entrant angle existed.

H. E. 27



418 Euclid's elements.



Exercises on Book XT.

1. Equal straight lines drawn to a plane from a point without
it have equal projections on that plane.

2. If S is the centre of the circle circumscribed about the triangle
ABC, and if SP is drawn perpendicular to the plane of the triangle,
shew that any point in SP is equidistant from the vertices of the
triangle.

3. Find the locus of points in space equidistant from three given
points,

4. From Example 2 deduce a practical method of drawing a
perpendicular from a given point to a plane, having given ruler,
compasses, and a straight rod longer than the required perpen-
dicular.

5. Give a geometrical construction for drawing a straight line
equally inclined to three straight lines which meet in a point, but are
not in the same plane.

6. In a gauche quadrilateral (that is, a quadrilateral whose sides
are not in the same plane) if the middle points of adjacent sides are
joined, the figure thus formed is a parallelogram.

7. AB and AC are two straight lines intersecting at right angles,
and from B a perpendicular BD is drawn to the plane in which they
are : shew that AD is perpendicular to AC.

8. If two intersecting planes are cut by two parallel planes, the
lines of section of the first pair with each of the second pair contain
equal angles.

9. If a straight line is parallel to a plane, shew that any plane
passing through the given stra,ight line will intersect the given plane
in a line of section which is parallel to the given line.

10. Two intersecting planes pass one through each of two
parallel straight lines ; shew that the common section of the planes
is parallel to the given lines.

11. If a straight line is parallel to each of two intersecting planes,
it is also parallel to the common section of the planes.

12. Through a given point in space draw a straight line to inter-
sect each of two given straight lines which are not in the same
plane.

13. If AB, BC, CD are straight lines not all in one plane, shew
that a plane which passes through the middle point of each one of them
is parallel both to AC and BD,

14. From a given point A a perpendicular AB is drawn to a
plane XY ; and a second perpendicular AE is drawn to a straight
line CD in the plane XY : shew that EB is perpendicular to CD.



EXERCISES ON BOOK XT. 419

15. From a point A two perpendiculars AP, AQ are drawn one
to each of two intersecting planes : shew that the common section of
these planes is perpendicular to the plane of AP, AQ.

16. From A, a point in one of two given intersecting planes,
AP is drawn perpendicular to the first plane, and AQ perpendicular
to the second : if these perpendiculars meet the second plane at P
and Q, shew that PQ is perpendicular to the common section of the
two planes.

17. A, B, C, D are four points not in one plane, shew that the
four angles of the gauche quadrilateral A BCD are together less than
four right angles.

18. OA, OB, OC are three straight lines drawn from a given
point O not in the same plane, and OX is another straight line
within the solid angle formed by OA, OB, OC : shew that

(i) the sum of the angles AOX, BOX, COX is greater than
half the sum of the angles AOB, BOC, COA.

(ii) the sum of the angles AOX, COX is less than the sum of
the angles AOB, COB.

(iii) the sum of the angles AOX, BOX, COX is less than the
sum of the angles AOB, BOC, COA.

19. OA, OB, OC are three straight lines forming a solid angle
at O, and OX bisects the plane angle AOB; shew that the angle
XOC is less than half the sum of the angles AOC, BOC.

20. If a point be equidistant from the angles of a right-angled
triangle and not in the plane of the triangle, the line joining it with
the middle point of the hypotenuse is perpendicular to the plane of
the triangle.

21. The angle which a straight line makes with its projection on
a. plane is less than that which it makes with any other straight line
which meets it in that plane.

22. Find a point in a given plane such that the sum of its
distances from two given points (not in the plane but on the same
side of it) may be a minimum.

23. If two straight lines in one plane are equally inclined to
another plane, they will be equally inclined to the common section
of these planes.

24. PA, PB, PC are three concurrent straight lines each of
which is at right angles to the other two: PX, PY, PZ are perpen-
diculars drawn from P to BO, CA, AB respectively. Shew that
XYZ is the pedal triangle of the triangle ABC.

25. PA, PB, PC are three concurrent straight lines each of which
is at right angles to the other two, and from P a perpendicular PO is
drawn to the plane of ABC : shew that O is the orthocentre of the
triangle ABC.

27-2



420



EUCLID'S ELEMENTS.



THEOBEMS AND EXAMPLES ON BOOK XT.



Definitions.

(i) Lines which are drawn on a plane, or through which a
plane may be made to pass, are said to be co-planar.

(ii) The projection of a line on a plane is the locus of the feet
of perpendiculars drawn from all points in the given line to the
plane.

Theorem 1. Tlie projection of a straight live on a plane is itself a
straight line.




Let AB be the given st. line, and XY the given plane.

From P, any point in AB, draw Pp perp. to the plane XY :

it is required to shew that the locus of 2> is a st. line.

From A and B draw Aa, Bb perp. to the plane XY.

Now since Aa, Pp, Bh are all perp. to the plane XY,

.-. they are par'. xi. 6,

And since these par'* all intersect AB,

.-. they are co-planar. xi. 7

the point p is in the common section of the planes Ah, XY ;
that is, p is in the st. line ab.
Butp is any point in the projection of AB,
.-. the projection of AB is the st. line ab. q.e.d



THEOREMS AND EXAMPLES ON BOOK XI.



421



Theorem 2. Draiv a perpendicular to each of two straight lines
which are not in the same plane. Prove that this perpendicular is the
shortest distance between the two lines.




Let AB and CD be the two straight lines, not in the same plane.

(i) It is required to draw a st. line perp. to each of them.

Through E, any point in AB, draw EF par^ to CD.

Let XY be the plane which passes through AB, EF.

From H, any point in CD, draw HK perp. to the plane XY. xi. 11.

And through K, draw KQ par^ to EF, cutting AB at Q.

Then KQ is also par^ to CD ; xi. 9.

and CD, HK, KQ are in one plane. xi. 7.

From Q, draw QP par' to HK to meet CD at P.
Then shall PQ be perp. to both AB and CD,
For, since HKis perp. to the plane XY, and PQ is par' to HK,

Constr.
:. PQ is perp. to the plane XY ; xi. 8.

.'. PQ is perp. to AB, which meets it in that plane. I)ef. 1.

For a similar reason PQ is perp. to QK,
.". PQ is also perp. to CD, which is par' to QK.

(ii) It is required to shew that PQ is the least of all st. lines
drawn from AB to CD.

Take HE, any other st. line drawn from AB to CD.
Then HE, being oblique to the plane XY is greater than the
perp. HK. p. 403, Ex. 1.

.•. HE is also greater than PQ. q.e.d.



422 EUCLID'S ELEMENTS.

Dkfinition. a parallelepiped is a solid figure bounded by three
pairs of parallel faces.

Theorem 3. (i) The faces of a parallelepiped are parallelograms,
of which those ichich are opposite are identically equal.

(ii) The four diagonals of a parallelepiped are concurrent and
bisect one another.




Let ABA'B' be a parPe*!, ^f ^hich ABCD, C'D'A'B' are opposite
faces.

(i) Then all the faces shall be par'"', and the opposite faces
shall be identically equal.

For since the planes DA', AD' are par', Bef.

and the plane DB meets them,
.-. the common sections AB and DC are par'. xi. 16.

Similarly AD and BC are par'.
.-. the fig. ABCD is a par™,
andAB = DC; alsoAD = BC. i. 34.

Similarly each of the faces of the parP«'' is a par*" ;
so that the edges AB, CD', B'A', DC are equal and par':
so also are the edges AD,, C'B', D'A', BC ; and likewise AC, BD',
CA', DB'.

Then in the opp. faces ABCD, C'D'A'B',

wehave AB = CD' and BC = D'A'; Proved.

and since AB, BC are respectively par' to CD', D'A' ,

.-. the Z ABC = the Z C'D'A'; xi. 10.

.-. the par*" ABCD = the par" C'D'A'B' identically. P. 64, Ex. 11.

(ii) The diagonals A A', BB', CC, DD' shall be concurrent and
bisect one another.

Join AC and A'C.



THEOREMS AND EXAMPLES ON BOOK XI. 423

Then since AC is equal and par' to A'C,
.-. the fig. ACA'C is a par'" ;
.-. its diagonals AA', CC bisect one another.
That is, A A' passes through O, the middle point of CC.
Similarly if BC and B'C were joined, the fig. BCB'C would be a
par™;

.'. the diagonals BB', CC bisect one another.
That is, BB' also passes through O the middle point of CC.
Similarly it may be shewn that DD' passes through, and is
bisected at, O. q.e.d.



Theorem 4. The straight lines which join the vertices of a tetra-
hedron to the centroids of the opposite faces are concurrent.




Let A BCD be a tetrahedron, and let g-^, g^, g^, g^ be the centroids
of the faces opposite respectively to A, B, C, D.

Then shall A^Ti, Bg^, Cg^, Dg^ be concurrent.
Take X the middle point of the edge CD ;
then g^ and </., must lie respectively in BX and AX,

so that BX = 3 . Xf7i , P. 105, Ex. 4.

and AX = 3. Xg^;
•■• 9i92 is pari to AB.
And A^jfj , Bg.2 must intersect one another, since they are both in
the plane of the "a AXB :

let them intersect at the point G.
Then by similar a^,, AG : G(/i = AB : g-^g^
= AX : X^2
= 3:1.
.*. B^2 cuts A^/i at a point G whose distance from^j = | . A^^.
Similarly it may be shewn that Cg.^ and D g^ cut Ar/j at the same
point ;

.-. these lines are concurrent. q.e.d.



424 EUCLID'S ELEMENTS.

Theorem 5. (i) If a pyramid is cut by planes drawn parallel to
its base, the sections are similar to the base.

(ii) The areas of such sections are in the duplicate ratio of their
perpendicular distances from the vertex,

S




A

Let SABCD be a pyramid, and abed the section formed by a
plane drawn par^ to the base A BCD.

(i) Then the figs. A BCD, abed shall be similar.
Because the planes abed, A BCD are par',

and the plane AB&a meets them,

.•. the common sections a6, AB are par'.

Similarly be is par' to BC; cd to CD; and da to DA.

And since ab, be are respectively par' to AB, BC,

.-. the z a6c = the z ABC. xi. 10.

Similarly the remaining angles of the fig. abed are equal to the
corresponding angles of the fig. A BCD.

And since the A" Sab, SAB are similar,
.-. ab : AB = S6 : SB

= bc : BC, for the a" Sbc, BCc are similar.
Or, ab : bc = AB : BC.

In like manner, 6c : cd =BC : CD.

And so on.
.-. the figs, abed, ABCD are equiangular, and have their sides
about the equal angles proportional.

/. they are similar.

(ii) From S draw Sa;X perp. to the planes abed, ABCD and
meeting them at x and X.

Then shall fig. abed : fig. ABCD = Sx2 : SX2.

Join ax, AX.

Then it is clear that the a' Sax, SAX are similar.

And the fig. abed : fig. ABCD = a62 : AB^ vi. 20.

= aS2 : AS2,

= Sx2 : SX\ Q.E.D.



DEFINITION. 425



Definition.

A polyhedron is regular when its faces are similar and equal
regular polygons.

Theorem 6. There cannot be more than five regular polyhedra.

This is proved by examining the number of ways in which it is
possible to form a solid angle out of the plane angles of various
regular polygons; bearing in mind that three plane angles at least
are required to form a solid angle, and the sum the plane angles
forming a solid angle is less than four right angles. xi. 21,

Suppose the faces of the 'regular polyhedron to be equilateral
triangles.

Then since each angle of an equilateral triangle is | of a right
angle, it follows that a solid angle may be formed (i) by three, (ii) by
four, or (iii) hy Jive such faces; for the sums of the plane angles
would be respectively (i) two right angles, (ii) f of a right angle,
(iii) -1/ of a right angle ;

that is, in all three cases the sum of the plane angles would be less
than four right angles.

^ But it is impossible to form a solid angle of six or more equilateral
triangles, for then the sum of the plane angles would be equal to, or
greater than four right angles.

Again, suppose that the faces of the polyhedron are squares.

(iv) Then it is clear that a solid angle could be formed of three,
but not more than three, of such faces.

Lastly, suppose the faces are regular pentagons.

(y) Then, since each angle of a regular pentagon is f of a right
angle, it follows that a solid angle may be formed of three such faces ;
but the sum of more than three angles of a regular pentagon is greater
than four right angles.

Further, since each angle of a regular hexagon is equal to | of a
right angle, it follows that no solid angle could be formed of such
faces ; for the sum of three angles of a hexagon is equal to four right
angles.

Similarly, no solid angle can be formed of the angles of a polygon
of more sides than six.

Thus there can be no more th&n five regular polyhedra.



42b-



EUCLID'S ELEMENTS.



Note on the Regular Polyhedra.

(i) The polyhedron of which each solid angle is formed by
three equilateral triangles is called a regular tetrahedron.



It has /our faces,

four vertices,
six edges.



(ii) The polyhedron of which each solid angle is formed by
four equilateral triangles is called a regular octahedron.






It has eight faces, six vertices, twelve edges.

(iii) The polyhedron of which each solid angle is formed by
Jive equilateral triangles is called a regular icosahedron.





It has twenty faces, twelve vertices, thirty edges.



NOTE ON THE REGULAR POLYHEDRA.



427



(iv) The regular polyhedron of which each solid angle is formed
by three squares is called a cutoe.



It has six faces,

eight vertices,
twelve




(v) The polyhedron of which each solid angle is formed by
three regular pentagons is called a regular dodecahedron.





It has twelve faces, twenty vertices, thirty edges.



428 Euclid's elements.

TuEOREsi 7. If F denote tlie number of faces, E of edges, and V of
vertices in any polyhedron, then toill

E + 2 = F + V.

Suppose the polyhedron to be formed by fitting together the faces
in succession : suppose also that E^ denotes the number of edges, and
V^ of vertices, when r faces have been placed in position, and that the
polyhedron has n faces when complete.

Now when one face is taken there are as many vertices as edges,
that is Ei = Vi.

The second face on being adjusted has two vertices and one edge in
common with the first; therefore by adding the second face we
increase the number of edges by one more than the number of
vertices; .*. E2-V2 = l.

Again, the third face on adjustment has three vertices and two
edges in common with the former two faces ; therefore on adding the
third face we once more increase the number of edges by one more
than the number of vertices ;

.-. E3-V3=2.

Similarly, when all the faces but one have been placed in position,
En-i-V„-i = 7i-2.

But in fitting on the last face we add no new edges nor vertices ;
•'• E=E„^i, V = V,i_i, and F=n.

So that E-V=F-2,
or, E + 2=F-i-V.

This is known as Eulefs Theorem.



Miscellaneous Examples on Solid Geometry.

1. The projections of parallel straight lines on any plane are
parallel.

2. If ah and cd are the projections of two parallel straight lines
AB, CD on any plane, shew that AB : CD = a6 : cd.

3. Draw two parallel planes one through each of two straight
lines which do not intersect and are not parallel.

4. If two straight lines do not intersect and are not parallel, on
what planes will their projections be parallel?

5. Find the locus of the middle point of a straight line of
constant length whose extremities lie one on each of two non-intersect-
ing straight lines.



MISCELLANEOUS EXAMPLES ON SOLID GEOMETRY. 429

0. Three points A, B, C are taken one on each of the conter-
minous edges of a cube: prove that the angles of the triangle ABC
are all acute.

7. If a parallelepiped is cut by a plane which intersects two pairs
of opposite faces, the common sections form a parallelogram.

8. The square on the diagonal of a rectangular parallelepiped is
equal to the sum of the squares on the three edges conterminous with
the diagonal.

9. The square on the diagonal of a cube is three times the square
on one of its edges,

10. The sum of the squares on the four diagonals of a parallele-
piped is equal to the sum of the squares on the twelve edges.

11. If a perpendicular is drawn from a vertex of a regular tetra-
hedron on its triangular base, shew that the foot of the perpendicular
will divide each median of the base in the ratio 2:1.

12. Prove that the perpendicular from the vertex of a regular
tetrahedron upon the opposite face is three times that dropped from
its foot upon any of the other faces.

13. If A P is the perpendicular drawn from the vertex of a regular
tetrahedron upon the opposite face, shew that

3AP2 = 2a2,

where a is the length of an edge of the tetrahedron.

14. The straight lines which join the middle points of opposite
edges of a tetrahedron are concurrent.

15. If a tetrahedron is cut by any plane parallel to two opposite
edges, the section will be a parallelogram.

16. Prove that the shortest distance between two opposite edges
of a regular tetrahedron is one half of the diagonal of the square on
an edge.

17. In a tetrahedron if two pairs of opposite edges are at right
angles, then the third pair will also be at right angles.

18. In a tetrahedron whose opposite edges are at right angles in
pairs, the four perpendiculars drawn from the vertices to the opposite
faces and the three shortest distances between opposite edges are
concurrent.

19. In a tetrahedron whose opposite edges are at right angles,
the sum of the squares on each pair of opposite edges is the same.

20. The sum of the squares on the edges of any tetrahedron is
four times the sum of the squares on the straight lines which join the
middle points of opposite edges.



430 EUCLID'S ELEMKNTK.

21. In any tetrahedron the plane which bisects a dihedral angle
divides the opposite edge into segments which are proportional to the
areas of the faces meeting at that edge.

22. If the angles at one vertex of a tetrahedron are all right
angles, and the opposite face is equilateral, shew that the sum of the
perpendiculars dropped from any point in this face upon the other
three faces is constant.

23. Shew that the polygons formed by cutting a prism by parallel
planes are equal.

24. Three straight lines in space OA, OB, OC, are mutually at
right angles, and their lengths are a, h, c : express the area of the
triangle ABC in its simplest form.

25. Find the diagonal of a regular octahedron in terms of one of
its edges.

26. Shew how to cut a cube by a plane so that the lines of section


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