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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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^ 3. Prove Simson's first Corollary by joining one vertex of the
rectilineal figure to each of the other vertices.

4. Find the magnitude of each angle of a regular polygon of
n sides.

5. If the alternate sides of any polygon be produced to meet, the
sum of the included angles, together with eight right angles, will
be equal to twice as many right angles as the figure has sides.



62 Euclid's elements.

Proposition IV^. Theorem.

The straight lines which join tite extremities of two equal
and parallel straight lines towards the same parts are them-
selves equal and parallel.




J^et AB and CD be equal and parallel straight lines;
and let them be joined towards the same parts by the
straight lines AC and BD:

then shall AC and BD be equal and parallel.

Construction. Join BC,

Proof. Then because AB and CD are parallel, and BC
meets them,

therefore the alternate angles ABC, BCD are equal, i. 29.

Now in tlie triangles ABC, DCB,
f AB is equal to DC, Hijp.

■n J and BC is common to both;

jalso the angle ABC is equal to the angle
\ DCB; Proved.

therefore the triangles are equal in all respects; I. 4.
so that the base AC is equal to the base DB,
and thfe angle ACB equal to the angle DBC;
but these are alternate angles ;
therefore AC and BD are parallel: i. 27.

and it has been shewn that they are also equal.

Q. E. D.

Definition. A Parallelogram is a four-sided figure



whose opposite sides are parallel.



o*



book 1. i'rop. 34, 63

Proposition 34. Theorem.

The opposite sides and angles of a parallelogra'ni are
equal to one another, and each diagonal bisects the jmrallelo-
gram.

3



\z\.



Let ACDB be a parallelogram, of which BC is a diagonal :
then shall the opposite sides and angles of the figure be
equal to one another; and the diagonal BC shall bisect it.

Proof. Because AB and CD are parallel, and BC meets
them,
therefore the alternate angles ABC, DCB are equal, i. 29.
Again, because AC and BD are parallel, and BC meets
them,

therefore the alternate angles ACB, DBC are equal, i. 29.
Hence in the triangles ABC, DCB,
rthe angle ABC is equal to the angle DCB,
p land the angle ACB is equal to the angle DBC;

jalso the side BC, which is adjacent to tlie equal
I angles, is common to both,
therefore the two triangles ABC, DCB are equal in all
respects; i. 26.

so that AB is equal to DC, and AC to DB;
and the angle BAC is equal to the angle CDB.
Also, because the angle ABC is equal to the angle DCB,
and the angle CBD equal to the angle BCA,
therefore the whole angle ABD is equal to the whole angle

DCA.
And since it has been shewn that the triangles ABC, DCB

are equal in all respects,
therefore the diagonal BC bisects the parallelogram ACDB.

Q.E.D.

[See note on next page.]



U4 KUCLID'.S KLEMENTri.

Note. To the proof which is here given Euclid added an applica-
tion of Proposition 4, with a view to shewing that the triangles ABC,
DCB are equal in area, and that therefore the diagonal BC bisects the
parallelogram. This equality of area is however sufficiently established
by the step which depends upon i. 2G. [See page 48.]



EXERCISES.

V

V' 1. Jj one angle oj a parallelogram is a right angle, all its angles are
right angles.

2. If the opposite sides of a quadrilateral are eqnal, the figure is a
parallelogram.

3. If the opposite angles of a quadrilateral are equal, the ^figure is
a parallelogram.

4. If a quadrilateral has all its sides equal and one angle a
right angle, all its angles are right angles; that is, all tlie angles of
a square are right angles.

5. The diagonals of a parallelogram bisect each other.

6. If the diagonals of a quadrilateral bisect each other, the figure
is a parallelogram.

7. If two opposite angles of a parallelogram are bisected by the
diagonal which joins them, the figure is equilateral.

8. If the diagonals of a parallelogram are equal, all its angles ai-e
right angles.

, 9. In a parallelogram which is not rectangular the diagonals are ?
unequal.

10. Any straight line drawn through the middle point of a diagonal
of a parallelogram and terminated by a pair of opposite sides, is
bisected at that point.

11. If tiDo parallelograms have two adjacent sides of one equal to
tico adjacent sides of the other, each to each, and one angle of one equal
to one angle of the other, the parallelograms are equal in all respects.

12. Two rectangles are equal if two adjacent sides of one are
equal to two adjacent sides of the other, each to each.

13. In a parallelogram the perpendiculars drawn from one pair of
opposite angles to the diagonal which joins the other pair are equal.

14. If ABCD is a parallelogram, and X, Y respectively the middle
points of tlie sides AD, BC ; shew that the figure AYCX is a parallelo-
gram.



>ll80ELLANK(>Urf EXKRCISES ON SECTIONS I. AND II. 05



MISCELLANEOUS EXERCISES ON SECTIONS T. AND II.

1. Shew that the construction in Proposition 2 may generally be
.performed in eight different ways. I'oint out the exceptional case.

2. The bisectors of two vertically opposite angles are in the same
.'straight line.

ii. In the figure of Proposition IG, if AF is joined, shew
(i) that AF is equal to BC; .

(ii) that the triangle ABC is equal to the triangle CFA in all V^
respects.

4. ABC is a triangle right-angled at B, and BC is produced to D: V
shew that the angle ACD is obtuse.

y. Shew that in any regular polygon of n sides each angle contains y^
right angles.

6. The angle contained by the bisectors of the angles at the base
of any triangle is equal to the vertical angle together with half the
sum of the base angles.

7. The angle contained by the bisectors of two exterior angles of
any triangle is equal to half the sum of the two corresponding interior
angles.

n

^ 8. If perpendiculars are drawn to two intersecting straight lines \
from any point between them, shew that the bisector of the angle
between the perpendiculars is parallel to (or coincident with) the
bisector of the angle between the given straight lines.

9. If two points P, Q. be taken in the equal sides of an isosceles
triangle ABC, so that BP is equal to CQ, shew that PQ is parallel to ^^'
BC.

10. ABC and DEF are two triangles, such that AB, BC are equal
and parallel to DE, EF, each to each; shew that AC is equal and
parallel to DF.

11. Prove the second Corollary to Prop. 32 by drawing through
any angular point lines parallel to all the sides.

12. If two sides of a quadrilateral are parallel, and the remaining
two sides equal but not parallel, shew that the opposite angles are
supplementary; also that the diagonals are equal.

H. E. 5



SECTION III.

THE AREAS OF PARALLELOGRAMS AND TRIANGLES.



Hitherto when two figures have been said to be equal, it has
been implied that they are identically equal, that is, equal in all
respects.

In Section III. of Euclid's first Book, we have to consider
the equality in area of parallelograms and triangles which are
not necessarily equal in all respects.

[The ultimate test of equality, as we have already seen, is afforded
by Axiom 8, which asserts that magnitudes which may he made to
coincide with one another are equal. Now figures which are not identi-
cally equal, cannot be made to coincide without first undergoing some
change of form : hence the method of direct superposition is unsuited
to the purposes of the present section.

We shall see however from Euclid's proof of Proposition 35, that
two figures which are not identically equal, may nevertheless be so
related to a third figure, that it is possible to infer the equality of
their areas.]

Definitions.

1 . The Altitude of a parallelogram with reference to a
given side as base, is the perpendicular distance between
the base and the opposite side.

2. The Altitude of a triangle with reference to a given
side as base, is the perpendicular distance of the opposite
vertex from the base.



book i. prop. 35. 6v

Proposition 35. Theorem.

Parallelograms on the same base, and hetimeii the sairie
parallels, are equal in area.



F A. D E F A E D F




Let the parallelograms ABCD, EBCF be on the same
base BC, and between the same parallels BC, AF :

then shall the parallelogram ABCD be equal in area to
the parallelogram EBCF.

Case I. If the sides of the given parallelograms, oppo-
site to the common base BC, are terminated at the same
point D :

then because each of the parallelograms is double of the

triangle BDC; i. 34.

therefore they are equal to one another. Ax. 6.

Case II. But if the sides AD, EF, opposite to the base

BC, are not terminated at the same point :

then because ABCD is a parallelogram,
therefore AD is equal to the opposite side BC; i. 34.
and for a similar reason, EF is equal to BC ;

therefore AD is equal to EF. Ax. 1.

Hence the whole, or remainder, EA is equal to the whole,
or remainder, FD.

Then in the triangles FDC, EAB,

C FD is equal to EA, Proved.

^ ^ J and DC is equal to the opposite side AB, i. 34.

jalso the exterior angle FDC is equal to the interior

\ opposite angle EAB, I. 29.

therefore the triangle FDC is equal to the triangle EAB. i. 4.

From the whole figure ABCF take the triangle FDC ;

and from the same figure take the equal triangle EAB ;

then the remainders are equal ; Ax. 3.

that is, the parallelogram ABCD is e£ua,l to the parallelo-
gram EBCF.




68 euclid'8 elements.



Proposition 3G. Theorem.



Paralleloyrams on equal bases, and between the same
parallels^ are equal in area.




Let ABCD, EFGH be parallelograms on equal bases BC,
FG, and between the same parallels AH, BG :
then shall the parallelogram ABCD be equal to the paral-
lelogram EFGH.

Construction. Join BE, OH.

Proof. Then because BC is equal to FG ; Hyp-

and FG is equal to the opposite side EH; i. 34.

therefore BC is equal to EH : Ax. 1.

and they are also parallel; Hyp.

therefore BE and CH, which join them towards the same
parts, are also equal and parallel. i. 33.

Therefore EBCH is a parallelogram. Def. 26.
Xow the parallelogram ABCD is equal to EBCH ;
for they are on the same base BC, and between the same
parallels BC, AH. i. 35.

Also the parallelogram EFGH is equal to EBCH;
for they are on the same base EH, and between the same
parallels EH, BG. i. 35.

Therefore the parallelogram ABCD is equal to the paral-
lelogram EFGH. Ax. 1.

Q. E. D.

From the last two Propositions we infer that :

(i) A parallelogram is equal in area to a rectarigle of equal
base and equal altitude.

(ii) Parallelograms on equal bases and of equal altitudes are
equal m area.



BOOK I. PROP. 37. 69

(iii) Of two parallelograms of equal altitudes, that is tJie greater
which has the greater base ; and of two parallelograms
on equal bases, that is the greater which has the greater
altitude.



Pko POSITION 37, Theohem.

Triangles on the same base, and between the same paral-
lels, are equal in area.




Let the triangles ABC, DBC be upon tlie same base BC,
and between the same parallels BC, AD.
Then shall the triangle ABC be equal to the triangle DBC.

Construction. Through B draw BE parallel to CA, to

meet DA produced in E; i. 31.

through C draw CF parallel to BD, to meet AD produced in F.

Proof. Then, by construction, each of the ligures EBCA,
, DBCF is a parallelogram. Def. 26.

And EBCA is equal to DBCF;
for they are on the same base BC, and between the same
parallels BC, EF. i. 35.

And the triangle ABC is half of the parallelogram EBCA,

for the diagonal AB bisects it. i. 34.

Also the triangle DBC is half of the parallelogram DBCF,
for the diagonal DC bisects it. i, 34.

But the halves of equal things are equal ; Ax. 7.
therefore the triangle ABC is equal to the triangle DBC.

Q.K.D.

[I'or Exercises see page 73. j



70 Euclid's elements.



Proposition 38. Theorem.

Triangles on equal bases, and between the same parallels,
are equal in area.




Let the triangles ABC, DEF be on equal bases BC, EF,
and between the same parallels BF, AD :
then shall the triangle ABC be equal to the triangle DEF.

Construction. Through B draw BG parallel to CA, to
meet DA produced in G; i. 31.

through F draw FH parallel to ED, to meet AD produced in H.

Proof. Then, by construction, each of the figures G BCA,
DEFH is a parallelogram. Def. 26.

And GBCA is equal to DEFH ;

for they are on equal bases BC, EF, and between the same

parallels BF, GH. i. 36.

And the triangle ABC is half of the parallelogram GBCA,

for the diagonal AB bisects it. i. 34.

Also the triangle DEF is half the parallelogram DEFH,

for the diagonal DF bisects it. i. 34.

But the halves of equal things are equal: Ax. 7.
therefore the triangle ABC is equal to the triangle DEF.

Q.E.D.

From this Proposition we infer that :

(i) Triangles on equal bases and of equal altitude are equal
in area.'

(ii) Of two triangles of tJie same altitude, that is the greater
which has the greater base : arid of two triangles on the same base,
or on equal bases, that is the greater which has the greater altitude.

[For Exercises see page 73.]



BOOK I. PROP. 39. 71



Proposition 39. Theorem.

Equal triangles on the same base, and oti the same side
of it, are het'wee7i the same parallels.




Let the triangles ABC, DBC which stand on the same
base BC, and on the same side of it, be equal in area :
then shall they be between the same parallels ;
that is, if AD be joined, AD shall be parallel to BC.

Construction. For if AD be not parallel to BC,

if possible, through A draw AE parallel to BC, i. 31,
meeting BD, or BD produced, in E.
Join EC.
Proof. Now the triangle ABC is equal to the triangle EBC,
for they are on the same base BC, and between the same
parallels BC, AE. i. 37.

But the triangle ABC is equal to the triangle DBC; Hyp.
therefore also the triangle DBC is equal to the triangle EBC;
the whole equal to the part ; which is impossible.
Therefore AE is not parallel to BC.
Similarly it can be shewn that no other straight line
through A, except AD, is parallel to BC.

Therefore AD is parallel to BC.

Q.E.D.

From this Proposition it follows that :

Equal triangles on the same base have equal altitudes.

[For Exercises see page 73.]



KLX'LIOS KI.KMENTS.



Proposition 40. Theorem.



Equal triangles, on equal bases in the same straight line,
and on the same side of it, are between the same parallels.




Let the triangles ABC, DEF wliich stand on equal bases
BC, EF, in the same straight line BF, and on the same side
of it, be equal in area :

then shall they be between the same parallels ;
that is, if AD be joined, AD si rail be parallel to BF.
Construction. For if AD be not parallel to BF,

if possible, through A draw AG parallel to BF, i. 31.
meeting ED, or ED produced, in G.
Join GF.

Proof. Now the triangle ABC is equal to the triangle GEF,
for they are on equal bases BC, EF, and between the
same parallels BF, AG. i. 38.

But the triangle ABC is equal to the triangle DEF: IIi/p.
therefore also the triangle DEF is equal to the triangle GEF :
the whole equal to the part; which is impossible.
Therefore AG is not parallel to BF.
Similarly it can be shewn that no other straight Hue
through A, except AD, is parallel to BF.

Therefore AD is parallel to BF.

From this Proposition it follows that :

(i) Equal triangles on equal bases hcvce equal altitudes.
(ii) Equal triangles of equal altitudes have equal bases.



EXEKCltiKS ON I'KOPlS. 37 — 40.



J^XERCISES OX PROPOSITIONS 37—40.

Definition. Each of tlie three straight lines wliich join
the angular points of a triangle to tlie middle points of the
opposite sides is called a Median of the triangle.

ON Prop. 37.

1. If, in the figure of Prop. 37, AC and BD intersect in K, shew that
(i) the triangles A KB, DKC are equal in area. "^^ '

(ii) the quadrilaterals EBKA, FCKD are equal.

2. In the figure of i. 16, shew that the triangles ABC, FBC are
c(|ual in area.

3. On the base of a given triangle construct a second triangle, \
equal in area to the first, and having its vertex in a given straight i
line. '^j

4. Describe an isosceles triangle equal in area to a given triangle
and standing on the same base.

ON Piior. 38.

5. .-1 triangle is divided by each of its medians into two parts of
equal area.

6. A parallelogram is divided by its diagonals into four triangles
of equal area.

7. ABC is a triangle, and its base BC is bisected at X; if Y '\
be any point in the median AX, shew that the triangles ABY, ACY are
equal in area.

8. In AC, a diagonal of the parallelogram A BCD, any point X is
taken, and XB, XD are drawn: shew that the triangle BAX is equal _■)
to the triangle DAX.

9. If two triangles have two sides of one respectively equal to two
sides of the other, and the angles contained by those sides supplement-
(irij, the triangles arc equal in area. ^ '

ON Pkop. 39.

10. Tlie straight line which joins the middle points of two sides of
a triangle is parallel to the third side.

11. If two straight lines AB, CD intersect in O, so that the triangle
AOC is equal to the triangle DOB, shew that AD and CB are parallel.

ON Piior. 40.

12. Deduce Prop. 40 from Prop. 39 by joining AE, AF in the
figure of page 72.



euclid's elements.
Pkoposition 41. Theorem.



If a parallelogram and a triangle he on tlin same base
and between the same j^^^rallels, the jyai'aUelogram shall be
double of the triangle.




Let the parallelogram ABCD, and the triangle EBC be
upon the same base BC, and between the same parallels
BC, AE :

then shall the parallelogram ABCD bo double of the triangle
EBC.
Construction. Join AC.

Proof. Then the triangle ABC is equal to the triangle EBC,
for they are on the same base BC, and between the same
parallels BC, AE. i. 37.

But the parallelogram ABCD is double of the triangle ABC,
for the diagonal AC bisects the parallelogram. 1. 34.

Therefore the parallelogram ABCD is also double of the
triangle EBC. q.e.d.



EXERCISES.

1. ABCD is a parallelogram, and X, Y are the middle points of
the sides AD, BC; if Z is any point in XY, or XY produced, shew-
that the triangle AZB is one quarter of the parallelogram ABCD.

2. Describe a right-angled isosceles triangle equal to a given square.

3. If ABCD is a parallelogram, and XY any points in DC and AD
respectively: shew that the triangles AXB, BYC are equal in area.

4. ABCD is a parallelogram, and P is any point within it; shew
that the sum of the triangles PAB, PCD is equal to half the paral-



lelogram.



TROP. 42.



Proposition 42. Problem.



To describe a jjarallelogram that shall he equal to a given
triangle, and have one of its angles equal to a given angle.



A F G




Let ABC be the given triangle, and D the given angle.
It is required to describe a parallelogram equal to ABC, and
having one of its angles equal to D.

Construction. Bisect BC at E. i. 10,

At E in CE, make the angle CEF equal to D ; i. 23.
through A draw AFG parallel to EC ; i. 31.

and through C draw CG parallel to EF.
Then FECG shall be the parallelogram required.
Join AE.

Froof. Now the triangles ABE, A EC are equal,
for they are on equal bases BE, EC, and between the same
parallels ; i. 38;

therefore the triangle ABC is double of the triangle A EC.
But FECG is a parallelogram by construction; Def. 2G.
and it is double of the triangle A EC,
for they are on the same base EC, and between the same
parallels EC and AG. i. 41.

Therefore the parallelogram FECG is equal to the triangle

ABC;
and it has one of its angles CEF equal to the given angle D.

Q. E. F.



EXERCISES.



1. Describe a parallelogram equal to a given square standing on
the same base, and having an angle equal to half a right angle.

2. Describe a rhombus equal to a given parallelogram and stand-
ing on the same base. When does the construction fail ?



70 EUCLlDfci ELEMENTS.

Definitiox. If in the diagonal of a parallelogram any
point is taken, and straight lines are drawn through it
parallel to the sides of the parallelogram; then of the four
parallelograms into which the whole figure is divided, tlie
two through which the diagonal passes are called Paral-
lelograms about that diagonal, and the other two, whicli
with these make up the whole figure, are called the
complements of the parallelograms about the diagonal.

Thus iu the figure given below, AEKH, KGCF are parallelograms
about the diagonal AC; and HKFD, EBGK are the complements of
those parallelograms.

Note. A parallelogram is often named by two letters only, these
being placed at opposite angular points.



PROPOSITION 43. Theorem.

27ie coinplemenls of the parallelograms about the diayonal
of any parallelogram, are equal to one another.




Let ABCD be a parallelogram, and KD, KB the comple-
ments of the parallelograms EH, GF about the diagonal AC:
then shall the complement BK be equal to the comple-
ment KD.
Proof. Because EH is a parallelogram, and AK its diagonal,
tli«refore the triangle AEK is equal to the triangle AHK. i. 34.
For a similar reason the triangle KGC. is equal to the

triangle KFC.
Hence the triangles AEK, KGC are together equal to the
triandes AHK, KFC.



BOOK I. PROr. 44. I I

But the whole triangle ABC is equal to the wliole triangle
ADC, for AC bisects the parallelogram ABCD ; i. 34.

therefore the remainder, the complement BK, is equal to the
remainder, the complement KD, Q.e.d.

EXERCISES.

In the figure of Prop. 43, ijvove that

(i) The parallelogram ED is equal to the parallelogram BH.

(ii) If KB, KD are joined, the triangle AKB is equal to the
triangle AKD.



Pi<oposiTioN 44. Problem.

To a given straight line to apply a jjarallelogratn vjhich
shall he equal to a given triangle, and have one of its angles
equal to a given angle.




Let AB be the given straight line, C the given triangle,
and D the given angle.

It is required to apply to the straight line AB a paral-
lelogram equal to the triangle C, and having an angle equal
to the angle D.

Construction. On AB produced describe a parallelogram
BEFG equal to the triangle C, and having the angle EBG
equal to the angle D; i. 22 and i. 42^.

through A draw AH parallel to BG or EF, to meet FG pro-
duced in H. I, 31.
Join HB.

* This step of the construction is effected by first describing on A B
produced a triangle whose sides are respectively equal to those of the
triangle C (i. 22) ; and by then making a parallelogram equal to the
triangle so drawn, and having an angle equal to D (i. 42).



EUCLUrS KI.EMENTH.




Then because AH and EF are parallel, and HF meets them,
therefore the angles AHF, HFE are together equal to two
right angles : I. 29.

hence the angles BHF, HFE are together less than two

right angles ;
therefore HB and FE will meet if produced towards B
and E. ' Ax. 12.

Produce them to meet at K.
Through K draw KL parallel to EA or FH; i. 31.
and produce HA, GB to meet KL in the points L and M.
Then shall BL be the parallelogram required.

Proof. Now FHLK is a parallelogram, Constr.

and LB, BF are the complements of the parallelograms
about the diagonal HK:

therefore LB is equal to BF. i. 43.

But the triangle C is equal to BF; Constr.

therefore LB is equal to the triangle C.
And because the angle GBE is equal to the vertically oppo-
site angle ABM, i. 15.
. and is likewise equal to the angle D ; Constr.
therefore the angle ABM is equal to the angle D.
Therefore the parallelogram LB, which is applied to the
straight line AB, is equal to the triangle C, and has the
angle ABM equal to the angle D. q.e.f.



BOOK I. TROP. 45. 79



Proposition 45. Problem.

To describe a 'parallelogram, equal to a given rectilineal
figure, and having an angle equal to a given angle.




H M

Let A BCD be the given rectilineal figure, and E the
given ansfle.

It is required to describe a parallelogram equal to A BCD,
and having an angle equal to E.
Suppose the given rectilineal figure to be a quadrilateral.

Construction. Join BD.

Describe the parallelogram FH equal to the triangle ABD,

and having the angle FKH equal to the angle E. I. 42.

To GH apply the parallelogram GM, equal to the triangle

DBC, and having the angle GHM' equal to E. i. 44.

Then shall FKML be the parallelogram required.

Proof. Because each of the angles GHM, FKH is equal to E,
therefore the angle FKH is equal to the angle GHM.
To each of these equals add the angle G H K ;
then the angles FKH, GHK are together equal to the angles
GHM, GHK.
But since FK, GH are jjarallel, and KH meets them,
therefore the angles FKH, GHK are together equal to two
right angles : i. 29.

therefore also the angles GHM, GHK are together equal to
two right angles :



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