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therefore KH, HM are in the same straight line. i. 14.



;UfMl)S Kl.KMKN



F .G L




H M



Again, l)ecause KM, FG are parallel, and HG meets tlieni,
therefore the alternate angles MHG, HGF are equal : i. 29

to each of these equals acid the angle HGL ;
then the angles MHG, HGL are together equal to the angles
HGF, HGL.

But because HM, GL are parallel, and HG meets them,
therefore the angles MHG, HGL are together equal to
two right angles: i. 29.

therefore also the angles HGF, HGL are together equal to
two right angles :

therefore FG, GL are in the same straight line. I. 14.

And boc.iuse KF and ML are each parallel to HG, Constr.

therefore KF is parallel to ML; i. 30.

and KM, FL are parallel ; Constr.

therefore FKML is a parallelogram. Def. 26.

And because the parallelogram FH is equal to tJie triangle

ABD, Constr.

and the parallelogram GM to the triangle DBG; Constr.

tlierefore the whole parallelogram FKML is equal to the

whole figure ABOD ;

and it has the angle FKM equal to the angle E.

By a series of similar steps, a parallelogram may be
constructed equal to a rectilineal figure of more than four
sides. Q.E.F.



BOOK I. PROP. 46.



81



Proposition 46. Problem.

To describe a square on a g'ive7i straight line.

' Cl



B

Let AB be the given straight line :
it is required to describe a square on AB.

Constr. From A draw AC at right angles to AB ; i. 11.
and make AD equal to AB. i. 13.

Through D draw DE parallel to AB; i. 31.

and tlirough B draw BE parallel to AD, meeting DE in E.
Then shall ADEB be a square.

Proof. For, by construction, ADEB is a parallelogram :
therefore AB is equal to DE, and AD to BE. I. 34.
But AD is equal to AB ; Constr.

therefore the four straight lines AB, AD, DE, EB are equal
to one another;

that is, the figure ADEB is equilateral.
Again, since AB, DE are parallel, and AD meets them,
therefore the angles BAD, ADE are together equal to two
right angles ; i. 29.

but the angle BAD is a right angle ; Constr.

therefore also the angle ADE is a right angle.
And the opposite angles of a parallelogram are equal ; i. 34.
therefore each of the angles DEB, EBA is a right angle :
that is the figure ADEB is rectangular.
Hence it is a square, and it is described on AB.

Q.E.P.

Corollary. If one angle of a yarallelogram is a right
angle, all its angles are right angles.

H. e. 6



82



EUCLID'S ELEMENTS,



Proposition 47. Theorem.

In a right-angled triangle the square described on ilm
hypotenuse is equal to the sum of the squares described on
tits otlter two sides.




Let ABC be a right-angled triangle, having the angle
BAG a right angle :

then shall the square described on the hypotenuse BC be
equal to the sum of the squares described on BA, AC.

Construction. On BC describe the square BDEC; i. 46.
and on BA, AC describe the squares BAGF, ACKH.

Through A draw AL parallel to BD or CE; I. 31.
and join AD, FC.

Proof. Then because each ol the angles BAC, BAG is a
right angle,

therefore CA and AG are in the same straight line. i. 14.

Now the angle CBD is equal to the angle FBA,

for each of them is a right angle.

Add to each the angle ABC :

then the whole angle ABD is equal to the whole angle FBC.



BOOK I. PROP. 47. 83

Then in the triangles ABD, FBC,

[ AB is equal to FB,

Because - and BD is equal to BC,

[also the angle ABD is equal to the angle FBC ;
therefore the triangle ABD is equal to the triangle FBC. 1.4.

Now the parallelogram BL is double of the triangle ABD,
for they are on the same base BD, and between the same
parallels BD, AL. I. 41.

And the square GB is double of the triangle FBC,

for they are on the same base FB, and between the same

parallels FB, GC. i. 41.

But doubles of equals are equal : Ax. 6.

therefore the parallelogram BL is equal to the square GB.

In a similar way, by joining AE, BK, it can be shewn
that the parallelogram CL is equal to the square CH.
Therefore the whole square BE is equal to the sum of the

squares GB, HC :
that is, the square described on the liypotenuse BC is equal

to the sum of the squares described on the two sides

BA, AC. Q.E.D,

Note. It is not necessary to the proof of this Proposition that
the three squares should be described external to the triangle ABC;
and since each square may be drawn either toicards or awa^j frovi the
triangle, it may be shewn that there are 2x2x2, or eight, possible
constructions.



EXERCISES.

1. In the figure of this Proposition, shew that

(i) If BG, CH are joined, these straight lines are parallel;

(ii) The points F, A, K are in one straight line;

(iii) FC and AD are at right angles to one another; i^

(iv) If GH, KE, FD are joined, the triangle GAH is equal
to the given triangle in all respects ; and the triangles
FBD, KCE are each equal in area to the triangle ABC.

[See Ex. 9, p. 73.]



84



EUCLID'S EL?:MENT8.



2. Ou the sides AB, AC of any triaugle ABC, squares ABFG,
ACKH are described both toward the triangle, or both on the side
remote from it: shew that the straight lines BH and CG are equal.

3. On the sides of any triangle ABC, equilateral triangles BCX,
CAY, ABZ are described, all externally, or all towards the triangle:
shew that AX, BY, CZ are all equal.

4. The square described on the diagonal of a given square, is
double of the given square.

5. ABC is an equilateral triangle, and AX is the perpendicular
drawn from A to BC : sheio that the square on AX is three times the
square on BX.

6. Describe a square equal to the sum of two given squares.

7. From the vertex A of a triangle ABC, AX is drawn perpendi-
cular to the hase : shew that the difference of the squares on the sides
AB and AC, is equal to the difference of the squares on BX and CX,
the segments of the base.

8. If from any point O within a triangle ABC, perpendiculars
OX, OY, OZ are drawn to the sides BC, CA, AB respectively; shew
that the sum of the squares on the segments AZ, BX, CY is equal to
the sum of the squares on the segments AY, CX, BZ.



Proposition 47. Alternative Proof.
C




Let CAB be a right-angled triangle, having the angle at A a right
angle :

then shall the square on the hypotenuse BC be equal to the sum of
the squares on BA, AC,



BOOK I. PROP. 47. 85

On A B describe the square A BFG. i. 46.

From FG and GA cut off respectively FD and GK, each equal

to AC. I. 3.

On GK describe the square GK EH : i. 46.

then HG and GF are in the same straight line. i. 14.

Join CE, ED, DB.

It will first be shewn that the figure CEDB is the square on CB.

Now CA is equal to KG ; add to each AK:

therefore OK is equal to AG.

Similarly DH is equal to GF:

hence the four lines BA, CK, DH, BF are all equal.

Then in the triangles BAG, CKE,

/ BA is equal to OK, Proved.

Because ^ ^^^ ^^ ^^ equal to KE; Constr.

' ^ j also the contained angle BAG is equal to the contained

' angle CKE, being right angles ;

therefore the triangles BAG, CKE are equal in all resi^ects. i. 4,

Similarly the four triangles BAG, CKE, DHE, BFD may be shewn

to be equal in all respects.

Therefore the four straight lines BC, CE, ED, DB are all equal;

that is, the figure CEDB is equilateral.

Again the angle CBA is equal to the angle DBF ; Proved.

add to each the angle ABD :

then the angle CBD is equal to the angle ABF :

therefore the angle CBD is a right angle.
Hence the figure CEDB is the square on BC. l)ef. 28.

And EHGK is equal to the square on AC. Comtr.

Now the square on CEDB is made up of the two triangles BAG, CKE,

and the rectilineal figure AKEDB ;
therefore the square CEDB is equal to the triangles EHD, DFB

together with the same rectilineal figure ;

but these make up the squares EHGK, AGFB:
hence the square CEDB is equal to the sura of the squares EHGK,

AGFB:
that is, the square on the hypotenuse BC is equal to the sum of the

squares on the two sides CA, AB. q. e. d.



Obs. The following properties of a square, though not
formally enunciated by Euclid, are employed in subsequent
proofs. [See i. 48.]

(i) 27ie squares on equal straiyltt Hues are equal.
(ii) Equal .squares stand tc2Jon equal straight lines.



86 euclid's elements.

Proposition 48. Theorem.

If the square described on one side of a triangle he equal
to the sum of the squares described on the oilier two sides, then
the angle contained by these two sides shall be a right angle.




Let ABC be a triangle ; and let the square described on

BC be equal to the sum of the squares described on BA, AC :

then shall the angle BAC be a right angle.

Construction. From A draw AD at right angles to AC; 1. 11.

and make AD equal to AB. i. .3.

Join DC.

Proof Then, because AD is equal to AB, Constr.

therefore the square on AD is equal to the square on AB.

To each of these add the square on CA;

then the sum of the squares on CA, AD is equal to the sum

of the squares on CA, AB.

But, because the angle DAC is a right angle, Constr.

therefore the square on DC is equal to the sum of the

squares on CA, AD. i. 47.

And, by hypothesis, the square on BC is equal to the sum

of the squares on CA, AB;

therefore the square on DC is equal to the square on BC :

therefore also the side DC is equal to the side BC.

Then in the triangles DAC, BAC,

/ DA is equal to BA, Constr.

-5 I and AC is common to both;

iJecause ^^^^^ ^j^^ ^j^.^^ ^.^^ ^^ .^ ^^^^^ ^^ ^^^^ ^^^.^^ ^.^^

[ BC ; Proved.

therefore the angle DAC is equal to the angle BAC. i. 8.
But DAC is a right angle ; Constr.

therefore also BAC is a right angle. Q. k. d.



THEOREMS AND EXAMPLES ON BOOK I.



INTRODUCTORY.

HINTS TOWARDS THE SOLUTION OF GEOMETRICAL EXERCISES.
ANALYSIS. SYNTHESIS.

It is commonly found that exercises in Pure Geometry present
to a beginner far more difficulty than examples in any other
branch of Elementary Mathematics. This seems to be due to
the following causes.

(i) The main Propositions in the text of Euclid must be not
merely understood, but thoroughly digested, before the exercises
depending upon them can be successfully attempted.

(ii) The variety of such exercises is practically unlimited;
and it is impossible to lay down for their treatment any definite
methods, such as the student has been accustomed to find in the
rules of Elementary Arithmetic and Algebra.

(iii) The arrangement of Euclid's Propositions, though per-
haps the most convincing of all forms of argument, affords in
most cases little clue as to the way in which the proof or con-
struction was discovered.

Euclid's propositions are arranged synthetically : that is
to say, they start from the hypothesis or data ; they next pro-
ceed to a construction in accordance with postulates, and pro-
blems already solved ; then by successive steps based on known
theorems, they finally establish the result indicated by the enun-
ciation.

Thus Geometrical Synthesis is a building up of known results,
in order to obtain a neio result.

But as this is not the way in which constructions or proofs
are usually discovered, we draw the attention of the student to
the following hints.

Begin by assuming the result it is desired to establish ; then
by working backwards, trace the consequences of the assumption,
and try to ascertain its dependence on some simpler theorem
which is already known to be true, or on some condition which
suggests the necessary construction. If this attempt is suc-
cessful, the steps of the argument may in general be re-arranged
in reverse order, and the construction and proof presented in a
synthetic form.



88



Euclid's elements.



This viuravelling of the conditions of a proposition in order
to trace it back to some earher principle on which it depends,
is called geometrical analysis : it is the natural way of attack-
ing most exercises of a more difficult type, and it is especially
adapted to the solution of problems.

These directions are so general that they cannot be said to
amount to a method: all that can be claimed for Geometrical
Analysis is that it furnishes a mode of searchiiig for a
suggestion^ and its success will necessarily depend on the skill
and ingenuity with which it is employed : these may be exi)ected
to come with experience, but a thorough grasp of the chief l^ro-
positions of Euclid is essential to attaining them.

The practical application of these hints is illustrated by the
following examples.

1. Construct an isosceles triaiujle having given the hose, and the
sum of one of the equal sides and the perpendicular drawn from the
vertex to the base.




Let AB be the given base, and K the sura of one side and the
l^erpendicular dx-awn from the vertex to the base.

Analysis. Suppose ABC to be the required triangle.

From C draw CX perpendicular to AB :

then AB is bisected at X. i. 2G.

Now if we produce XC to H, making XH equal to K,
it follows that CH =CA ;
and if AH is joined,
we notice that the angle CAH = the angle CHA. i. 5.

Now the straight lines XH and AH can be drawn before the position
of C is known ;

Hence we have the following construction, which we arrange
synthetically.



THEOREMS AND EXAMPLES ON BOOK I.



89



Synthesis. Bisect AB at X :

from X draw XH perpendicular to AB, making XH equal to K.
Join AH.
At the point A in HA, make the angle HAC equal to the angle
AHX ; and join CB.

Then ACB shall be the triangle required.

First the triangle is isosceles, for AC = BC. i. 4.

Again, since the angle HAC = the angle AHC, Constr.

.-. HC = AC. I. G.

To each add CX ;
then the sum of AC, CX = the sum of HC, CX
= HX.
That is, the sum of AC, CX= K. q. e. f.

2. To divide a given straight line so that the sqnare on one part
may he double of the square on the other.



Let AB be the given straight line.

Analysis. Suppose AB to be divided as required at X : that is,
suppose the square on AX to be double of the square on XB.

Now we remember that in an isosceles right-angled triangle, the
square on the hypotenuse is double of the square on either of the
equal sides.

This suggests to us to draw BC perpendicular to AB, and to make
BC equal to BX,

Join XC.
Then the square on XC is double of the square on XB, i. 47.
.-. XC = AX.
And when we join AC, we notice that

the angle XAC = the angle XCA, i. 5.

Hence the exterior angle CXB is double of the angle XAC. i. 32.
But the angle CXB is half of a right angle : i. 32.

.-. the angle XAC is one-fourth of a right angle.

This supplies tlie clue to the following construction :—



90 EUCLID'S l.l.KilKNT.N.

Synthesis. From B draw BD perpendicular to AB ;
and from A draw AC, makiny BAG one-fourth of a right angle.
From C, the intersection of AC and BD, draw CX, making the angle
ACX equal to the angle BAC. i. 23.

Then AB shall be divided as required at X.
For since the angle XCA = the angle XAC,

.-. XA = XC. 1.6.

And because the angle BXC = the sum of the angles BAC, ACX, i. 32.
.-. the angle BXC is half a right angle;
and the angle at B is a right angle ;
therefore the angle BCX is half a right angle ; i. 32.

therefore the angle BXC = the angle BCX ;
.-. BX = BC.
Hence the square on XC is double of the square on XB : i. 47.
that is, the square on AX is double of the square on XB. q.e.f.



I. 0\ THE IDENTICAL EQUALITY OF TRIANGLES.

See Propositions 4, 8, 26.

1. If in a triangle the perpendicular from the vertex on the base
bisects the base, then the triangle is isosceles.

2. If the bisector of the verti<Jal angle of a triangle is also per-
pendicular to the base, the triangle is isosceles.

3. If the bisector of the vertical angle of a triangle also bisects
the base, the triangle is isosceles.

[Produce the bisector, and complete the construction after the
manner of i. 16.]

4. If in a triangle a pair of straight lines drawn from the ex-
tremities of the base, making equal angles with the sides, are equal, the
triangle is isosceles.

5. If in a triangle the perpendiculars drawn from the extremities
of the base to the opposite sides are equal, the triangle is isosceles.

6. Two triangles ABC, ABD on the same base AB, and on opposite
sides of it, are such that AC is equal to AD, and BC is equal to BD :
shew that the line joining the points C and D is perpendicular to AB.

7. If from the extremities of the base of an isosceles triangle per- j
pendiculars are drawn to the opposite sides, shew that the straight '
line joining the vertex to the intersection of these perpendiculars bisects
the vertical angle.



THEOREMS AND EXAMPLES ON BOOK I.



91



8. ABC is a triangle in which the vertical angle BAC is bisected
by the straight line AX : from B draw BD perpendicular to AX, and
produce it to meet AC, or AC produced, in E; then shew that BD is
equal to DE.

9. In a quadrilateral ABCD, AB is equal to AD, and BC is equal
to DC : shew that the diagonal AC bisects each of the angles which it
joins.

10. In a quadrilateral ABCD the opposite sides AD, BC are equal,
and also the diagonals AC, BD are equal: if AC and BD intersect at
K, shew that each of the triangles AKB, DKC is isosceles.

11. If one angle of a triangle be equal to the sum of the other two,
the greatest side is double of the distance of its middle point from the
opposite angle.

12. Two right-angled triangles which have their hypotenuses equal,
and one side of one equal to one side of the other, are identically equal.




Let ABC, DEF be two a ^ right-angled at B and E, having AC
equal to DF, and AB equal to DE :

then shall the A ^ be identically equal.
For apply the a ABC to the a DEF, so that A may fall on D,
and AB along DE ; and so that C may fall on the side of DE remote
from F.

Let C be the point on which C falls.
Then since AB=DE,
.'. B must fall on E ;
so that DEC represents the A ABC in its new position.
Now each of the / » DEF, DEC is a rt. l ;
.-. EF and EC are in one st. line.
Then in the a CDF,
because DF=DC,
.-. the Z DFC=:the z DCF.
Hence in the two A^ DEF, DEC,
( the Z DEF = the z DEC, being rt. l'
Because] and the Z DFE=the z DCE;

( also the side D E is common to both ;

.-. the A" DEF, DEC are equal in all respects; i. 20.

that is, the a^ DEF, ABC are equal in all respects. q.k.d.



Hyp.
I. 14.



Praised.



92 Euclid's elements.

13. If two triangles have two sides of the one equal to two sides of
the other, each to each, and have likewise the angles opposite to one pair
of equal sides equal, then the angles opposite to the other pair of equal
sides are either equal or supplementary, and in the former case the tri-
angles are equal in all respects.




Let ABC, DEF be two AS

Imviug tlie side AB equal to the side DE,

the side AC equal to the side DF,

also the Z ABC equal to the z DEF:

then shall the Z ^ ACB, DFE be either equal or supplementary,

and in the former case, the a^ shall be equal in all respects.

Apply the a ABC to the a DEF,
no that A may fall on D, and AB along DE ;

then because AB = DE, J^'/P-

.'. B will fall on E :
and because the Z ABC = the z DEF, Ifyp.

. .'. BC will fall along EF:
Then must C fall on F, or in EF, or EF produced.

If C falls on F,

the A ** coincide, and thei'efore are identically equal ;

so that the z ACB = the z DFE.

But if C falls in EF, or EF produced, as at C :
then DEC represents the a ABC in its new position.

Then because DF = AC, JiyP'

.'. DF = DC',
.-. the z DC'F = the z DFC. i. 5.

But the z >* DFC, DFE are supplementary; i. 13.

.". the Z * DCF, DFE are supplementary :
that is, the z =* ACB, DFE are supplementary. q.e.d.

Three cases of this theorem deserve special attention.

It has been proved that if the angles ACB, DFE are not equal,
they are supjilementary :

And we know that of angles which are supplementary and un-
equal, one must be acute and the other obtuse.



THEOREMS AND EXAMPLES ON BOOK I. 93

Corollaries. Hence if in addition to the hypothesis of this
theorem it is given

(i) That the angles AC B, D F E, opposite to the two equal sides
AB, DE are both acute, both obtuse, or if one of them
is a right angle,

it follows that these angles are equal,
and therefore that the triangles are equal in all respects.

(ii) That the two given angles are right angles or obtuse
angles, it follows that the angles ACB, DFE must be
both acute, and therefore equal, by (i) :
so that the triangles are equal in all respects.

(iii) That in each triangle the side opposite the given angle
is not less than the other given side ; that is, if AC and
DP are not less than AB and DE respectively, then
the angles ACB, DFE cannot be greater than the angles
ABC, DEF, respectively;

therefore the angles ACB, DFE, are both acute ;
hence, as above, they are equal ;
and the triangles ABC, DEF are equal in all respects.



II. ON INEQUALITIES.

See Propositions 16, 17, 18, 19, 20, 21, 24, 25.



1. In a triangle ABC, if AC is not greater than AB, shew that
any straight line drawn through the vertex A, and terminated by the
base BC, is less than AB.

2. ABC is a triangle, and the vertical angle BAC is bisected by a
straight line which meets the base BC in X ; shew that BA is greater
than BX, and CA greater than CX. Hence obtain a proof of i. 20. .

3. llie perpendicular is the shortest straight line that can be
drawn from a given point to a given straight line ; and of others, that
xchich is nearer to the perpendicular is less than the more remote ; and
tiiw, and only tioo equal straight lines can be drawn from the given
point to the given straight line, one on each side of the perpendicular.

4. The sum of the distances of any point from the three angular
points of a triangle is greater than half its perimeter.

5. The sum of the distances of any point within a triangle from
its angular points is less than the perimeter of the triangle.



94 Euclid's elements.

6. The perimeter of a quadrilateral ib greater than the 8uni of its
diagonals.

7. The sum of the diagonals of a quadrilateral is less than the
sum of the four straight lines drawn from the angular points to any
given point. Prove this, and point out the exceptional case.

8. In a triangle any Uco sides are together greater than twice the
median which bisects the remaining side. [See Def. p. 73,]

[Produce the median, and complete the construction after the
manner of i. 16.]

9. In any triangle the sum of the medians is less than the peri-
meter.

10. In a triangle an angle is acute, obtuse, or a right angle,
according as the median drawn from it is greater than, less than, or
equal to half the opposite side. [See Ex. 4, p. 59.]

11. The diagonals of a rhombus are unequal.

12. If the vertical angle of a triangle is contained by unequal
sides, and if from the vertex the median and the bisector of the angle
are draicn, then the median lies within tlie angle contained by the
bisector and the longer side.

Let ABC be a a, in which AB is greater
than AC ; let AX be the median drawn from
A, and AP the bisector of the vertical
ZBAC:

then shall AX lie between AP and AB.
Produce AX to K, making XK equal to
AX. Join KC.

Then the a' BXA, CXK may be shewn

to be equal in all respects; i. 4.

hence BA = CK,and the z BAX = the z CKX.

But since BA is greater than AC, IlTjp.

:. CK is greater than AC;

.-. the Z CAK is greater than the z CKA: i. 18.

that is, the z CAX is greater than the z BAX :
.-. the Z CAX must be more than half the vert, z BAG ;

hence AX lies within the angle BAP. q.e.d.

13. If two sides of a triangle are unequal, and if from their point
of intersection three straight lines are drawn, namely the bisector of the
vertical angle, the median, and the perpendicular to the base, the first
is intermediate in position and magnitude to the other two.




THEOREMS AND EXAMPLES ON BOOK I. 95

III. ON PARALLELS.

See Propositions 27 — 31.

1. If a straight line meets two parallel straight lines, and the
two interior angles on the same side are bisected; shew that the
bisectors meet at right angles, [i. 29, i. 32.]

2. The straight lines drawn from any point in the bisector of
an angle parallel to the arms of the angle, and terminated by them,
are equal ; and the resulting figure is a rhombus.

3. AB and CD are two straight lines intersecting at D, and the
adjacent angles so formed are bisected: if through any point X in
DC a straight line YXZ be drawn parallel to AB and meeting the
bisectors in Y and Z, shew that XY is equal to XZ.

4. If two straight lines are parallel to two other straight lines,
each to each; and if the angles contained by each pair are bisected;
shew that the bisecting lines are parallel.

5. The middle point of any straight line which meets two parallel
straight lines, and is terminated by them, is equidistant from the



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