Euclid.

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book i. prop. 26. 45

Proposition 26. Theorem.

If two triangles have two angles of the one equal to two
angles of the other, each to each, and a side of one equal
to a side of the other, these sides beiyig either adjacent to the
equal angles, or opposite to equal angles in each ; then shall
the triangles he equal iri all respects.

Case T. When the equal sides are adjacent to the equal
ani^les in the two triangles.

A D

Let ABCj DEF be two triangles, which have the angles
ABC, ACB, equal to the two angles DEF, DFE, each to each;
and the side BC equal to the side EF :

then shall the triangle ABC be equal to the triangle DEF
in all respects ;

that is, AB shall be equal to DE, and AC to DF,
and the angle BAC shall be equal to the angle EDF.
For if AB be not equal to DE, one must be greater than
the other. If possible, let AB be greater than DE.
Construction. From BA cut off BG equal to ED, i. 3.

and join GC.
Proof Then in the two triangles GBC, DEF,

iGB is equal to DE, Constr.

and BC to EF, Hyp.

also the contained angle GBC is equal to the
contained angle DEF; ^J^yp-

therefore the triangles are equal in all respects ; i. 4.
so that the angle GCB is equal to the angle DFE.
But the angle ACB is equal to the angle DFE ; Hyp.
therefore also the angle GCB is equal to the angle ACB ; Ax.\.
the part equal to the whole, which is impossible.

46

JiUCLIDW ELEMENTS.

Because

Tlierefore AB is not unequal to DE,

that is, AB is equal to DE,
Hence in the triangles ABC, DEF,

AB is equal to DE, Proved.

and BC is equal to EF ; Hyp-

ABC is equal to the
V contained angle DEF ; J^yp-

therefore the triangles are equal in all respects : i. 4.
so that the side AC is equal to the side DF ;

and the angle BAC to the angle EDF. q e.d.

also the contained angle

Case If. When the equal sides are opposite to equal

angles in the two triangles.

Let ABC, DEF be two triangles Avhich have the angles

ABC, ACB equal to the angles DEF, DFE, each to each,

and the side AB equal to the side DE :

then shall the triangles ABC, DEF be equal in all respects ;

that is, BC shall be equal to EF, and AC to DF,

and the angle BAC shall be equal to the angle EDF.

BOOK I. PROP. 26. â–  47

For if BC be not equal to EF, one must be greater than
the other. If possible, let BC be greater than EF.

Construction. From BC cut off BH equal to EF, i. 3.

and join AH.

Proof. Then in the triangles ABH, DEF,

{AB is equal to DE, J^yp-

and BH to EF, Constr.

also the contained angle ABH is equal to the
contained angle DEF ; ^-^2//^-

therefore the triangles are equal in all respects, i. 4.
so that the angle AHB is equal to the angle DFE.
But the angle DFE is equal to the angle ACB ; JJ^l/]^-
therefore tlie angle AHB is equal to the angle ACB ; Ax. 1.
that is, an exterior angle of the triangle ACH is equal to
an interior opposite angle ; which is impossible. i. 1 6.
Therefore BC is not unequal to EF,

that is, BC is equal to EF.
Hence in the triangles ABC, DEF,
( AB is equal to DE, J^^P-

and BC is equal to EF ; Proved.

also the contained angle ABC is equal to the
contained angle DEF ; f^yp-

therefore the triangles are equal in all respects; i. 4.
so that the side AC is equal to the side DF,

i Because

and the ansjle BAC to the angle EDF.

Q.E.D.

Corollary. In both cases of this Projjosition it is seen
that the triangles may he made to coincide with one another;
and they are therefore equal in area.

48 Euclid's elements.

ON THE IDENTICAL EQUALITY OF TRIANGLES.

At the close of the first section of Book I., it is worth while
to call special attention to those Propositions (viz. Props, 4, 8, 26)
which deal with the identical equality of two triangles.

The results of these Propositions may be summarized thus :

Two triangles are equal to one another in all respects, when
the following parts in each are equal, each to each.

1. Two sides, and the included angle. Prop. 4.

2. The three sides. Prop. 8, Cor.

3. (a) Two angles, and the adjacent side. \

(b) Two angles, and the side opposite one of >- Prop. 26.
them. )

From this the beginner will perhaps' surmise that two tri-
angles may be shewn to be equal in all respects, when they have
three parts equal, each to each; but to this statement two obvious

(i) When in two triangles the three angles of one are equal
to the three angles of the other, each to each, it does not
necessarily follow that the triangles are equal in all respects.

(ii) When in two triangles two sides of the one arc equal
to two sides of the other, each to each, and one angle equal to
one angle, these not being the angles included by the equal sides ;
the triangles are not necessarily equal in all respects.

In these cases a further condition must be added to the*
hypothesis, before we can assert the identical equality of the
two triangles.

[See Theorems and Exercises on Book I., Ex. 13, Page 92.]

We observe that in each of the three cases already proved
of identical equality in two triangles, namely in Propositions 4,
8, 26, it is shewn that the triangles may be made to coincide
with one another ; so that they are equal in area, as in all
other respects. Euclid however restricted himself to the use of
Prop. 4, when he required to deduce the equality in area of two
triangles from the equality of certain of their parts.

This restriction has been abandoned in the present text-book.
[See note to Prop. 34.]

EXERCISES ON PROPS. 12 â€” 26. 49

EXERCISES ON PROPOSITIONS 12 â€” 26.

1. If BX and CY, the bisectors of the angles at the base BC of afi
isosceles triangle ABC, meet the opposite sides in X and Y ; shew that
the triangles YBC, XCB are equal in all respects.

2. Shew that the perpendiculars drawn from the extremities of
the base of an isosceles triangle to the opposite sides are equal.

3. Any point on the bisector of an angle is equidistant from the
arms of the angle.

4. Through O, the middle point of a straight line AB, any straight
line is drawn, and perpendiculars AX and BY are dropped upon it from
A and B : shew that AX is equal to BY.

5. If the bisector of the vertical angle of a triangle is at right
angles to the base, the triangle is isosceles.

6. The perpendicular is the shortest straight line that can he
draion from a given point to a given straight line ; and of others, Jiat
which is nearer to the perpendicular is less than the more remote; and
two, and only two equal straight lines can be draion from the given
point to the given straight line, one on each side of the perpendicular.

7. From two given points on the same side of a given straight line,
draic two straight lines, which shall meet in the given straight line

make equal angles icith it.

Let AB be the given straight line,
and P, Ql the given points.

It is required to draw from P and Q.
to a point in AB, two straight lines
that shall be equally inclined to AB.

Construction. From P draw PH
pfrpendicular to AB: produce PH to

P', making HP' equal to PH. Draw QP', meeting AB in K.
PK.

Then PK, QK shall be the required lines. [Supply the proof.]

8. In a given straight line find a point which is equidistant from
two given intersecting straight lines. In what case is this impossible?

9. Through a given point draw a straight line such that the per-
pendiculars drawn to it from two given points may be equal.

In what case is this impossible?

H. E. 4

50 EUCLID'S ELEMENTS.

SECTION 11.

PARALLEL STRAIGHT LINES AND PARALLELOGRAMS.

Definition. Parallel straight Irnes are such as, being
in the same plane, do not meet however far they are pro-
duced in both directions.

When two straight lines A B, CD are met by a third straight
line EF, eight angles are formed, to which
for the sake of distinction particular
names are given.

1, 2, 7, 8 are called exterior angles,
3, 4, 5, 6 are called interior angles,
4 3^nd 6 are said to be alternate angles ;
so also the angles 3 and 5 are alternate to
one anotlier.

Of the angles 2 and 6, 2 is referred to as the exterior angle,
and 6 as the interior opposite angle on the same side of EF.

2 and 6 are sometimes called corresponding angles.

So also, 1 and 5, 7 and 3, 8 and 4 are corresponding angles.

Euclid's treatment of parallel straight lines is based upon his
twelfth Axiom, which we here repeat.

Axiom 12. If a straight line cut two straight lines so
as to make the two interior angles on the same side of
it together less than two right angles, these straight lines,
being continually produced, will at length meet on that
side on which are the angles which are together less than
two right angles.

Thus in the figure given above, if the two angles 3 and 6 are
together less than two right angles, it is asserted that AB and
CD will meet towards B and D.

This Axiom is used to establish i. 29 : some remarks upon it
will be found in a note on that Proposition.

BOOK T. PRor. 27. 51

Propositiox 27. Theorem.

If a idraight line, falling on two other straight lines, make
the alternate angles equal to one another, then the straight
lines shall be parallel.

Let the straight line EF cut the two straight lines AB,
CD at G and H, so as to make the alternate angles AGH,
G H D equal to one another :

then shall AB and CD be parallel.

Proof For if AB and CD be not parallel,
they will meet, if produced, either towards B and D, or to-
wards A and C.
If possible, let AB and CD, when produced, meet towards B

and D, at the point K.
Then KG H is a triangle, of which one side KG is produced

to A:

therefore the exterior angle AG H is greater than the interior

opposite angle GHK. i. 16.

But the angle AG H is equal to the angle GHK: Hyj)-

Jience the angles AGH and GHK are both equal and unequal;

which is impossible.
Therefore AB and CD cannot meet when produced towaixls.

B and D.
Similarly it may be shewn that they cannot meet towards
A and C:

therefore they are parallel. q.e. d-

4â€”2.

b-2 EUCLID .S ELEMENTS.

f

Tko POSITION 28. Theorem.

// a straight line, fallifig on two otiier straight lineSj
iiiake an exterior arigle equal to the interior opposite angle
on the same side of the line; of if it make the interior
ajigles on the same side together equal to two right angles,
then the two straight lines shall he parallel.

Let the straight line EF ciit the two straiglit Hues AB,
CD in G and H : and

First, let the exterior angle EG B be equal to the interior
opposite angle G H D :

then shall AB and CD be parallel.
Proof Because the angle EGB is equal to the angle GHD;
and because the angle EG B is also equal to the vertically op-
posite angle AG H ; 1.15.
therefore the angle AGH is equal to the angle GHD;
but these are alternate angles;
therefore AB and CD are parallel. i. 27.

Q. E. D.

Secondly, let the two interior angles BGH, GHD be to-
gether equal to two right angles :

then shall AB and CD be parallel.

Proof Because the angles BGH, GHD arc together equal

to two right angles; ih/P-

and because the adjacent angles BGH, AGH are also together

equal to two right angles ; 1.13.

therefore the angles BGH, AGH are together equal to the

two angles BGH, GHD.

From these equals take the common angle BGH :
then the remaining angle AGH is equal to the remaining
angle GHD: and these are alternate angles;

therefore AB and CD are parallel. i. 27.

Q. E. I).

BOOK 1. PROP. 29. 53

%

Proposition 29. Theorem.

If a straight line fall on two imrallel straight lines, then it
shall make the alternate angles equal to one another , and the
exterior angle equal to the interior opposite angle on the
same side; and also the two interior angles on the same
side equal to two right angles.

Let- the straight line EF. fall on the parallel straight
lines AB, CD:

then (i) the alternate angles AGH, GHD shall be equal to
one another;
(ii) the exterior angle EG B shall be equal to the interior

opposite angle GHD;
(iii) the two interior angles BGH, GHD shall be together

â– s equal to two right angles

'roof (i) For if the angle AGH be not equal to the angle
GHD, one of them must be greater than the other.
If possible, let the angle AGH be greater than the angle
GHD;

add to each the angle BGH :
then the angles AGH, BGH are together greater than the

angles BGH, GHD.
But the adjacent angles AGH, BGH are together equal to
two right angles ; i. 13.

therefore the angles BGH, GHD are together less than two
right angles;

therefore AB and CD meet towards B and D. Ax. 12.
But they never meet, since they are parallel. li^J^-
Therefore the angle AG H is not unequal to the angle GHD:
that is, the alternate angles AGH, GHD are equal.

{Over)

54 Euclid's elements.

(ii) Again, because the angle AGH is equal to tlie verti-
cally opposite angle EGB; L 15.
and because the angle AGH is equal to the angle GHD;

Proved.

therefore the exterior angle EGB is equal to the interior op-
posite angle GHD.

(iii) Lastly, the angle EGB is equal to the angle GHD;

Proved.
add to each the angle EG H ;
then the angles EG B, BG H are together equal to the angles

BGH, GHD.
But the adjacent angles EGB, BGH are together equal to
two right angles: I. 13.

therefore also the two interior angles BGH, GHD are to-
gether equal to two right angles. q.e.d.

EXERCISES ox PROPOSITIONS 27, 28, 29.

1. Two straight lines AB, CD bisect one another at O: shew that
the straight Hnes joining AC and BD are parallel. [i. 27.J

2. Straight lines which are perpendicular to the same straight line
are parallel to one another. [i. 27 or i. 28.J

3. If a straight line meet two or more parallel straight lines, aiul is
perpendicular to one of them, it is also perpendicular to all the others.

[I. 29.]

4. If two straight lines are parallel to two other straight lines, each
to each, then the angles contained by the first pair are equal respectivehj ^
to the angles contained hy the second pair. [i. 29.] '

BOOK I. PROP. 29. 55

XoTE ox THE Twelfth Axiom.

It must bo admitted that Euclid's twelfth Axiom is un-
satisfactory as the basis of a theory of parallel straight lines.
It cannot be regarded as either simple or self-evident, and it
therefore falls short of the essential characteristics of an axiom :
nor is the difficulty entirely removed by considering it as a cor-
roUary to Proposition 17, of which it is the converse.

Many substitutes have been proposed ; but we need only notice
here the system which has met with most general approval.

This system rests on the following hypothesis, which is put
forward as a fundamental Axiom.

Axio:m. Two intersecting straight lines cannot he both parallel
to a third straight line.

This statement is known as Playfair's Axiom ; and though
it is not altogether free from objection, it is recommended as
both simpler and more fundamental than that employed by

Propositions 27 and 28 having been proved in the usual way,
the first part of Proposition 29 is then given thus.

Proposition 29. [Alternative Proof.J

Jf a straight line fall on two parallel straight lines, then it
shall make the alternate angles eqnal.

Let the straight line EF meet the two
parallel straight lines AB, CD, at G
and H :
then shall the alternate angles AGH,

G H D be equal.
For if the angle AG H is not equal to the

angle GHD:
at G in the straight line HG make the
angle HGP equal to the angle GHD,
and alternate to it. i. 28.

Then PG and CD are parallel, i. 27.
But AB and CD are parallel: Hyp.
therefore the two intersecting straight lines AG, PG are both parallel
to CD:

which is impossible. Flay fair's Axiom.

Therefore the angle AG H is not unequal to the angle GHD,
that is, the alternate angles AGH, GHD are equal, q.e.d.
The second and third parts of the Proposition may then be deduced
as in the text; and Euclid's Axiom 12 follows as a Corollary.

56 euclid's elements.

Proposition 30. Theorem.

Straight lines ivJiich are j)(^rallel to the same straight line
are jiaraUel to oiie another.

Z

C 7- D

p â€” ÂĄ^ â€” â– â–  Q

Let tlie straight lilies AB, CD be eacli })arallel to tlie
straight line PQ : .

then sliall AB and CD be parallel to one anotlier.

Construction. Draw any straight line EF cutting AB, CD,
and PQ in the points G, H, and K.

Proof. Then because AB and PQ are parallel, and EF
meets them,

therefore the angle AG K is equal to the alternate angle G KQ.

L 29.
And because CD and PQ are parallel, and EF meets them,
therefore . the exterior angle GHD is equal to the interior
opposite angle HKQ. L 29.

Therefore the angle AGH is equal to the angle GHD;
and these are alternate angles;
therefore AB and CD are parallel. i. 27.

Q.E.D.

Note. If PQ lies between AB and CD, the Proposition may be
estabhshed in a similar manner, though in this case it scarcely needs
proof; for it is inconceivable that two straight Knes, which do not
meet an intermediate straight line, should meet one another.

The truth of this Proposition may be readily deduced from
Playfair's Axiom, of which it is the converse.

For if AB and CD were not parallel, they would meet when pro-
duced. Then there would be two intersecting straight lines both
parallel to a third straight line : which is impossible.

Therefore AB and CD never meet; that is, they are parallel.

BOOK 1. PKor. 31. 57

Proposition 31. Problem.

To draw a straight line through a given point parallel
to a given straight li7ie.

1

Let A be the given point, and BC the given straight line.
It is required to draw through A a straight line parallel to
BC.

Construction. Jn BC take any point D; and join AD.
At the point A in DA, make the angle DAE equal to the
angle ADC, and alternate to it i. 23.

and produce EA to F.
Then shall EF be parallel to BC.

Proof. Because the straight line AD, meeting tlie two

equal; Constr.

therefore EF is parallel to BC; i. 27.

and it has been drawn through the given j^oint A.

Q, E. F.

EXERCISES.

1. Any straight line drawn parallel to the base of an isosceles
triangle makes equal angles with the sides.

2. If from any point in the bisector of an angle a straight line is
drawn parallel to either arm of the angle, the triangle thus formed is
isosceles.

3. From a given point draw a straight line that shall make with
a given straight line an angle equal to a given angle.

4. From X, a point in the base BC ol an isosceles triangle ABC, a
traight line is drawn at right angles to the base, cutting AB in Y, and

CA produced in Z : shew the triangle AYZ is isosceles.

5. If the straight line which bisects an exterior angle of a triangle
is parallel to the base, shew that the triangle is isosceles.

58, euciaivs elements.

Proposition 32. TheoreiM.

If a side of a triangle he jji'oducedj then the exterior
angle shall be equal to the sum of the two interior opposite
a)igles: also the three interior angles of a triangle are together
equal to two right angles.

i

Let ABC be a triangle, and let one of its sides BC be
produced to D :

then (i) the exterior angle ACD shall be equal to the sum
of the two interior opposite angles CAB, ABC;
(ii) the three interior angles ABC, BCA, CAB sliall
be together equal to two right angles.

Construction. Through C draw CE parallel to BA. i. 31.

Proof (i) Then because BA and CE are parallel, and AC

meets them,
therefore tlie angle ACE is equal to the alternate angle

CAB. I. 29.

Again, because BA and CE are parallel,. and BD meets them,
therefore the exterior angle ECD is equal to the interior

opposite angle ABC. I. 29.

Therefore the whole exterior angle ACD is equal to the

sum of the two interior opposite angles CAB, ABC.

(ii) Again, since the angle ACD is equal to the sum of
the angles CAB, ABC ; Proved.

to each of these equals add the angle BCA :
then the 'angles BCA, ACD are together equal to the thi'ee

angles BCA, CAB, ABC.
But the adjacent angles BCA, ACD are together equal to
two right anfjles; i. 13.

therefore also the angles BCA, CAB, ABC are together equal
to two r\a;ht angles. q. e. d.

BOOK I. PROP. 32. 59

From this Proposition we draw the foiloAving important
inferences.

1. If tico triangles have two angles of the one equal to two angles of
the other, each to each, then the third angle of the one is equal to the
third angle of the other. Â«

2. In any right-angled triangle the two acute angles are com-
plementary.

3. In a right-angled isosceles triangle each of the equal angles
is half a right angle.

4. If one angle of a triangle is. equal to the sum of the other two,
the triangle is right-angled.

5. The sum of the angles of any quadrilateral figure is equal to
four right angles.

6. Each angle of an equilateral triangle is two-thirds of a right
angle.

EXERCISES ON PROPOSITIOX 32

1. Prove that the three angles or a triangle are together etiual to
two right angles,

(i) hy drawing through the vertex a straight line parallel

to the base ;
(ii) by joining the verteis to any point in the base.

2. If the base of any triangle is produced both Avays, shew that
the sum of the two exterior angles diminished by the vertical angle is
equal to two right angles.

3. If two straight lines are perpendicular to two other straight
lines, each to each, the acute angle between the first pair is equal
to the acute angle between the second pair.

4. Every right-angled triangle is divided into tivo isosceles tri-
angles by a straight line draxvnfrom the right angle to the middle point
of the hypotenuse.

Hence the joining line is equal to half the hypotenuse*

I 5. Draw a straight line at right angles to a given finite straight
line from one of its extremities, without producing the given straight
line.

[Let AB be the given straight line. On AB describe any isosceles
triangle ACB. Produce BC to D, making CD equal to BC. Join

60 EUCLID 8 ELEMENTS.

6. Trisect a right angle. ^

7. The angle contained by the bisectors of the angles at the base
of an isosceles triangle is equal to an exterior angle formed by pro-
ducing the base.

8. The angle contained by the bisectors of two adjacent angles of
H quadrilateral is equal to half the sum of the remaining angles.

Tlie following theorems were added .-is corollaries to
Proposition 32 by Robert Simson.

Corollary 1. All the inteHor angles of any rectilineal
figure, iwith four Hght angles, J are together equal to twice as
many right angles as the figure has sides.

Let ABODE be any rectilineal figure.

Take F, any point within it,

and join F to each of the angular points of the figure.

Then the figure is divided into as many triangles as it has
sides.

And the three angles of each triangle are together equal
to two right angles. I. 32.

Hence all the angles of all the triangles are together equal
to twice as many right angles as the figure has sides.

But all the angles of all the triangles make up the in-
terior angles of the figure, together with the angles
at F;

and the angles at F are together equal to four right
angles: i. 15, Cor.

Therefore all the interior angles of tlie figure, with four
right angles, are together equal to twice as many right
angles as the figure has sides. Q. E. d.

BOOK I. riioi'. ?r2. 01

Corollary 2. If the sides of a rectilineal Jiyure, ichich
has no re-entrant angle, are iwoduced in order, theri all the ex-
terior angles so formed, are together equal to four right angles.

ft

For at each angular point of the figure, the interior angle
and the exterior angle are together equal to two right
angles. L 13.

Therefore all the interior angles, with all tlie exterior
angles, are together equal to twice as many right angles
as the figure has sides.

But all the interior angles, with four right angles, are to-
gether equal to twice as many right angles as the figure
has sides. i. 32, Cor. 1.

Therefore all the interior angles, with all the exterior
angles, are together equal to all the interior angles, with
four right an^jles.

Therefore the exterior angles are together equal to four
right angles. Q. e. d.

EXERCISES ON SIMSON S COROLLARIES.

UA polygon is said to be regular when it has all its sides and all its
angles equal.]

, 1. Express in terms of a right angle the magnitude of each angle
of (i) a regular hexagon, (ii) a regular octagon.

2. If one side of a regular hexagon is produced, shew that the ex-
terior angle is equal to the angle of an equilateral triangle.

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