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

. **(page 3 of 27)**

Online Library → Euclid → A text-book of Euclid's Elements : for the use of schools : Books I-VI and XI → online text (page 3 of 27)

Font size

BOOK T. PROP. 12. 27

EXERCISES ON PROPOSITIONS 1 TO 12.

1. Shew tbat the straight line which joins the vertex of an

isosceles triangle to the middle point of the base is perpendicular

to the base.

2. Shew that the straight lines which join the extremities of the

base of an isosceles triangle to the middle points of the opposite sides,

are equal to one another.

* 3. Two given points in the base of an isosceles triangle are equi-

distant from the extremities of the base : shew that they are also equi-

distant from the vertex.

4. If the opposite sides of a quadrilateral are equal, shew that the

opposite angles are also equal.

" 5. Any two isosceles triangles XAB, YAB stand on the same base

AB: shew that the angle XAY is equal to the angle XBY; and that

the angle AXY is equal to the angle BXY.

6. Shew that the opposite angles of a rhombus are bisected by the

diagonal which joins them.

' 7. Shew that the straight lines which bisect the base angles of an

isosceles triangle form with the base a triangle which is also isosceles.

8. ABC is an isosceles triangle having AB equal to AC ; and the

angles at B and C are bisected by straight lines which meet at O :

shew that OA bisects the angle BAC.

/ 9. â– Shew that the triangle formed by joining the middle points of

the sides of an equilateral triangle is also equilateral.

I 10. The equal sides BA, CA of an isosceles triangle BAC are pro-

duced beyond the vertex A to the points E and F, so that AE is equal

to AF; and FB, EC are joined: shew that FB is equal to EC.

11. Shew that the diagonals of a rhombus bisect one another at

right angles.

12. In the equal sides AB, AC of an isosceles triangle ABC two

points X and Y are taken, so that AX is equal to AY; and CX and BY

are drawn intersecting in O : shew that

(i) the triangle BOC is isosceles;

(ii) AC bisects the vertical angle BAC ;

(iii) AG, if produced, bisects BC at right angles.

13. Describe an isosceles triangle, having given the base and the

length of the perpendicular drawn from the vertex to the base.

14. In a given straight line find a point that is equidistant from

two given points.

In what case is this impossible ?

EUCLID'S ELEMENTS.

Proposition 13. Theorem.

If one straight line stand upon anotlier straight line,

tlien the adjacent angles shall he either tivo right angles, or

together equal to two right angles.

A E A

Let the straight line AB stand upon the straight line DC :

then the adjacent angles DBA, ABC shall be either two right

angles, or together equal to two right angles.

Case I. For if the angle DBA is equal to tlie angle ABC,

each of them is a right angle. Def. 7.

Case II. But if the angle DBA is not equal to the

angle ABC,

from B draw BE at right angles to CD. i. 11.

Proof. Now the angle DBA is made up of the two

angles DBE, EBA;

to each ofÂ» these equals add the angle ABC;

then the two angles DBA, ABC are together equal to the

three angles DBE, EBA, ABC. Ax. 2.

Again, the angle EBC is made up of the two angles EBA,

ABC;

to each of these equals add the angle DBE.

Then the two angles DBE, EBC are together equal to the

three angles DBE, EBA, ABC. Ax. 2.

But the two angles DBA, ABC have been shewn to be equal

to the same three angles ;

therefore tlie angles DBA, ABC are together equal to the

angles DBE, EBC. Ax. 1.

But the angles DBE, EBC are two right angles; Constr.

therefore the angles DBA, ABC are together equal to two

light angles. Q. e. d.

BOOK I. PROr. 13.

Definitions.

(i) The complement of an acute angle is its defect from

a right angle, that is, the angle by which it falls short of a right

angle.

Thus two angles are complementary, when their sum is a

right angle.

(ii) The supplement of an angle is its defect from two right

angles, that is, the angle by which it falls short of two right

angles.

Thus two angles are supplementary, when their sum is two

right angles.

Corollary. Arigles which are comide^nentary or supple-

mentary to the same angle are equal to one another.

EXERCISES.

1. If the two exterior angles formed by producing a side of a tri-

angle both ways are equal, shew that the triangle is isosceles.

2. The bisectors of the adjacent angles which one straight line

makes with another contain a right angle.

Note. Jn the adjoining figure AOB

is a given angle; and one of its arms AO

is produced to C: the adjacent angles

AOB, BOC are bisected by OX, OY.

Then OX and OY are called respect-

ively the internal and external bisectors

of the angle AOB.

Hence Exercise 2 may be thus enunciated :

The internal and external bisectors of an angle are at right angles

to one another.

3. Shew that the angles AOX and COY are complementary.

4. Shew that the angles BOX and COX are supplementary; and

also that the angles AOY and BOY are supplementary.

^ ' Euclid's elemei^ts.

v^J*ROPOSiTiON 14. Theorem.

Jj\ at a point hi <i straight Ihie, ttvo other stra'ujht lineSy

on opjiosite sides of it, make the adjacent angles togetlter

equal to ttvo right angles, then these two straight lines shall

he in one and the same straight liv.e.

At the point B in the straight line AB, let the two

straight lines BC, BD, on the opposite sides of AB, make

the adjacent angles ABC, ABD together equal to two right

angles :

then BD shall be in the same straight line with BC.

Proof. For if BD be not in the same straight line with BC,

if possible, let BE be in the same straight line with BC.

Then because AB meets the straight line CBE,

therefore the adjacent angles CBA, ABE are together equal

to two right angles. i. 13.

But the angles CBA, ABD are also together equal to two

right angles. Hyj).

Therefore the angles CBA, ABE are together equal to the

angles CBA, ABD. Ax. 11.

From each of these equals take the common angle CBA;

then the remaining angle ABE is equal to the remaining angle

ABD; the part equal to the whole; wliich is impossible.

Therefore BE is not in the same straight line with BC.

And in the same way it may be shewn that no other

line but BD can be in the same straight line with BC.

Therefore BD is in the same straight line with BC. q.e.d,

EXERCISE. y^

A BCD is a rhombus; and the diagonal AC is bisected at O. If O

is joined to the angular points B and D; shew that OB and CD are

in one straight line.

BOOK I. PROP. 15. .31

Ohs. When two straight lines intersect at a point, four

angles are formed; and any two of these angles which are not

adjacent, are said to be vertically opposite to one another.

Proposition 15. Theorem.

If two straight lines intersect dne'ctHother, theyi the verticallij

vpi^osite angles shall he equal.

A

Let the two straight lines AB, CD cut one another at

the point E :

then shall the angle A EC be equal to the angle DEB,

and the angle CEB to the angle A ED.

Proof. Because AE makes with CD the adjacent angles

CEA, AED,

therefore these angles are together equal to two right

angles. 1. 13.

Again, because DE makes with AB the adjacent angles AED,

DEB,

therefore these also are together equal to two right angles.

Therefore the angles CEA, AED are together equal to the

angles AED, DEB.

From each of these equals take the common angle AED;

then the remaining angle CEA is equal to the remaining

angle DEB. Ax. 3.

In a similar way it may be shewn that the angle CEB

is equal to the angle AED. q.e.d.

Corollary 1. From this it i^ manifest that, if two

straight liyies cut one another, tJie angles which they make

at the point where they cut, are together equal to four right

angles.

Corollary 2. Consequently, when ayiy number of straight

lines meet at a point, the sum of the angles made by con-

secutive Ihies is equal to four right angles.

. EUCLID'S ELEMENTS.

Proposition 16. Theorem.

If one aide of a triangle he j}roduced, then the exterior a^ujle

slmll he greater than either of the interior opposite amjles.

Because

Let ABC be a triangle, and let one side BC be produced

to D : then shall the exterior angle ACD be greater than

either of the interior opposite angles CBA, BAG.

Construction. Bisect AC at E : i. 10.

Join BE; and produce it to F, making EF equal to BE. l 3.

Join FC.

Proof. Then in the triangles AEB, CEF,

AE is equal to CE, Constr.

and EB to EF ; Constr.

also the angle AEB is equal to tlie vertically

opposite angle CEF; I. 15.

therefore the triangle AEB is equal to the triangle CEF in

all respects : i. 4.

so that the angle BAE is equal to the angle ECF.

But the angle ECD is greater than its part, the angle ECF;

therefore the angle ECD is greater than the angle BAE;

that is, the angle ACD is greater than the angle BAC.

In a similar way, if BC be bisected, and the side AC

produced to G, it may be shewn that the angle BCG is

greater than the angle ABC.

But the angle BCG is equal to the angle ACD: i. 15.

therefore also the angle ACD is greater than the angle ABC.

Q. E. D.

book i. prop. 17. 33

Proposition 17. Theorem.

Any two angles of a triangle are together less than two

right angles.

A k

B

Let ABC be a triangle; then shall any two of its angles, as

ABC, ACB, be together less than two riljht angles.

Construction. Produce the side BC to D.

Proof. Then because ACD is an exterior angle of the

triangle ABC,

therefore it is gieater than the interior opposite angle

ABC. I. 16.

To each of these add the angle ACB :

then the angles ACD, ACB are together greater than the

angles ABC, ACB. Ax. 4.

But the adjacent angles ACD, ACB are together equal to

two right angles. i. 13.

Therefore the angles ABC, ACB are together less than two

right angles.

Similarly it may be shewn that the angles BAC, ACB, as

also the angles CAB, ABC, are together less than two right

angles. q. e. d.

Note. It follows from this Proposition that every triangle must

have at least two acute angles : for if one angle is obtuse, or a right

angle, each of the other angles must be less than a right angle.

EXERCISES.

1. Enunciate this Proposition so as to shew that it is the converse l^

of Axiom 12.

2. If any side of a triangle is produced both ways, the exterior

angles so formed are together greater than two right angles. *

3. Shew how a proof of Proposition 17 may be obtained by

joining each vertex in turn to any point in the opposite side.

H. E. *)

34 EUCLID'S ELEMENTS.

Proposition 18. Theorem.

If one side of a triangle be greater than another, then

the angle opposite to the greater side shall be greater than

the angle opposite to the less.

Let ABC be a triangle, in wliich the side AC is greater

than the side AB :

then shall the angle ABC be greater than the angle ACB.

Construction. From AC, the greater, cut off a part AD equal

to AB. L 3.

Join BD,

Proof. Then in the triangle ABD,

because AB is equal to AD,

therefore the angle ABD is equal to the angle ADB. i. 5,

But the exterior angle ADB of the triangle BDC is

greater than the interior opposite angle DCB, that is,

greater than the angle ACB. i. 16.

TJierefore also the angle ABD is greater than the angle ACB;

still more then is the angle ABC greater than the angle

ACB. ^ Q.E.D.

. Euclid enunciated Proposition 18 as follows:

The greater side of every tricmgle has the greater angle

opposite to it.

[This form of enunciation is found to be a common source of diffi-

culty with beginners, who fail to distinguish what is assumed in it and

what is to be proved.]

[For Exercises see page 38.]

I

BOOK I. PROP. 19. 35

Proposition 19. Theorem.

If one angle of a triangle he greater than another^ then

the side opjjosite to the greater angle shall he greater than

the side opposite to the less.

I

Let ABC be a triangle in wliich the angle ABC is greater

tlian the angle ACB ;

then shall the side AC be greater than the side AB.

Proof For if AC be not greater than AB,

it must be either equal to, or less than AB.

But AC is not equal to AB,

for then the angle ABC would be equal to the angle ACB j i. 5.

but it is not. U^Vl^-

Neither is AC less than AB ;

for then the angle ABC would be less than the angle ACB ; I.18.

but it is not : HyP-

Therefore AC is neither equal to, nor less than AB.

That is, AC is greater than AB. q.e.d.

Note, The mode of demonstration used in this ProiDosition is

known as the Proof by Exhaustion. It is applicable to cases in which

one of certain mutually exclusive suppositions must necessarily be

true; and it consists in shewing the falsity of each of these supposi-

tions in turn loith one exception: hence the truth of the remaining

supposition is inferred.

Euclid enunciated Proposition 1 9 as follows :

The greater angle of every triangle is suhtended hy the

greater side, or, has the greater side opposite to it.

[For Exercises see page 38.]

3â€”2

36 Euclid's elements.

Proposition 20. Theorem.

Any two sides of a tt'iangle are together greater than the

third side.

B

Let ABC l)e a triangle: â€¢

then sliall any two of its sides be together greater tlian the

third side :

name]}', BA, AC, shall be greater than CB ;

AC, CB greater than BA ;

and CB, BA greater than AC.

Construction. Produce BA to the point D, making AD equal

to AC. 1. 3.

Join DC.

Proof. Then in the triangle ADC,

because AD is equal to AC, Constr.

therefore the angle ACD is equal to the angle ADC. i. 5.

But the angle BCD is greater than the angle ACD ; Ax. 9.

therefore also the angle BCD is greater than the angle ADC,

that is, than the angle BDC.

And in the triangle BCD,

because the angle BCD is greater than the angle BDC, Pr.

therefore the side BD is greater than the side CB. i. 19.

But BA and AC are together equal to BD ;

therefore BA and AC are together greater than CB.

Similarly it may be shewn

that AC, CB are together greater than BA ;

and CB, BA are together greater than AC. Q. E. D.

[For Exercises aee page 38.]

jioOK I. PROP. i^l. 37

Propositiox 21. Theorem.

If from the ends of a side of a triangle, there he drawn

two straight lines to a j^oint withiri the triangle, then these

straight lines shall be less than the other two sides of the

triangle, hut shall contain a greater angle.

I

Let ABC be a. triangle, aiid from B, C, the ends of tlie

. side BC, let the two straight lines BD, CD be drawn to

a point D wdthin the triangle :

then (i) BD and DC shall be together less than BA and AC ;

ii) the angle BDC shall be greater than the angle BAC.

onstruction. Pi-oduce BD to meet AC in E.

Proof (i) In the triangle BAE, the two sides BA, AE an^

together greater than the third side BE : i, 20.

to each of these add EC :

then BA, AC are together greater than BE, EC. Ax. 4.

Again, in the triangle DEC, the two sides DE, EC are to-

gether greater than DC : - i. 20.

to each of these add BD ;

then BE, EC are together greater than BD, DC.

But it has been shewn that BA, AC are together greater

than BE, EC :

still more tlien are BA, AC greater than BD, DC.

ii) Again, the exterior angle BDC of the triangle DEC is

greater than the interior opposite angle DEC ; i. 16.

and the exterior angle DEC of the triangle BAE is greater

than the interior opposite angle BAE, that is, than the

angle BAC ; i. 16.

still more then is the angle BDC greater than the angle BAC.

Q.E.D.

L th

%

38 Euclid's elements,

exercises

oil PUOPOSITIONS 18 AND 19.

1. The hypotenuse is the greatest side of a right-angled triangle.

2. If two angles of a triangle are equal to one another, the sides

also, which subtend the equal angles, are equal to one another. Prop. 6.

Prove this indirectly by using the result of Prop. 18.

3. BC, the base of an isosceles triangle ABC, is produced to any

point D ; shew that AD is greater than either of the equal sides.

4. If in a quadrilateral the greatest and least sides are opposite to

one another, then each of the angles adjacent to the least side is

greater than its opposite angle.

'â– 5. 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.

â– 6. ABC is a triangle, in which OB, GO bisect the angles ABC.

ACB respectively: shew that, if AB is greater than AC, then OB is

greater than OC.

OK Proposition 20.

!

7. The difference of any two sides of a triangle is less than

the third side.

8. In a quadrilateral, if two opposite sides which are not parallel

are produced to meet one another; shew that the perimeter of the

greater of the two triangles so fonned is greater than the perimeter of

the quadrilateral.

^. The sum of the distances of any point from the three angular

points of a triangle is greater than half its perimeter.

10. The perimeter of a quadrilateral is greater than the sum of its

diagonals.

11. Obtain a proof of Proposition 20 by bisecting an angle by a

straight line which meets the opposite side.

ox Proposition 21.

12. In Proposition 21 shew that the angle BDC is greater than

the angle BAC by joining AD, and producing it towards the base.

13. The sum of the distances of any point within a triangle from

its angular points is less than the perimeter of the triangle.

BOOK I. PROP. 22.

39

Proposition 22. Problem.

To describe a triangle having its sides equal to three

given straight lines, any two of which are together greater

than the third.

B

Let A, B, C be the three given straight lines, of which

any two are together greater than the third.

It is required to describe a triangle of which the sides

shall be equal to A, B, C.

Construction. Take a straight line DE terminated at tlie

point D, but unlimited towards E.

Make DF equal to A, FG equal to B, and GH equal to C. I. 3.

From centre F, with radius FD, describe the circle DLK.

From centre G with radius GH, describe the circle MHK,

cutting the former circle at K.

Join FK, GK.

Then shall the triangle KFG have its sides equal to the

tliree straight lines A, B, C.

Proof. Because F is the centre of the circle DLK,

therefore FK is equal to FD : Def 11.

but FD is equal to A ; Constr.

. therefore also FK is equal to A. Ax. 1.

Again, because G is the centre of the circle MHK,

therefore GK is equal to GH : Def. 11.

but G H is equal to C ; Constr.

therefore also GK is equal to C. Ax. 1.

And FG is equal to B. Constr.

Therefore the triangle KFG has its sides KF, FG, GK equal

respectively to the three given lines A, B, C. q.e.f.

40 EUCLIDS KLEilKNTS.

Â£XERCISÂ£.

Ou a given base describe a triangle, whose remaining sides shall be

equal to two given straight lines. Point out how the construction

fails, if any one of the three given lines is greater than the sum of

the other two.

Proposition 23. Problem.

At a given point in a given straigJtt line, to make an

angle equal to a given angle.

Let AB be the given straight line, and A the given point

in it; and let DCE be the given angle.

It is required to draw from A a straight line making

with AB an angle equal to the given angle DCE.

Construction. In CD, CE take any points D and E ;

and join DE.

From AB cut off AF equal to CD. I. 3.

On AF describe the triangle FAG, having the remaining

sides AG, GF equal respectively to CE, ED. i. 22.

Then shall the angle FAG be equal to the angle DCE.

Proof. For in the triangles FAG, DCE,

( FA is equal to DC, Constr.

Because < and AG is equal to CE; Constr.

(and the base FG is equal to the base DE : Constr.

therefore the angle FAG is equal to the angle DCE. i. 8.

That is, AG makes with AB, at the given point A, an angle

equal to the given angle DCE. q.e.f.

BOOK I. PROP. 24. 41

PROPOSITION 24.

Jf two triangles have two sides of the one equal to two

sides of the other, each to each, but the angle contained by

the two sides of one gi-eater than the angle contained by

the corresponding sides of the other ; then the base of that

which has the greater angle shall be greater than the base of the

other.

D'

Let ABC, DEF be two triangles, in wliicli the two sides

BA, AC are equal to the two sides ED, DF, each to each,

but the angle BAC greater than the angle EDF :

then shall the base BC be greater than the base EF.

' * Of the two sides DE, DF, let DE be that which is not

greater than DF.

Construction. At the point D, in the straight line ED,

and on the same side of it as DF, make the angle EDG

equal to the angle BAC. ]. 23.

Make DO equal to DF or AC; i. 3.

and join EG, GF.

Proof Then in the triangles BAC, EDG,

I' BA is equal to ED, HyjJ.

-D and AC is equal to DG, Constr.

I also the contained angle BAC is equal to the

\ contained angle EDG ; Constr.

Therefore the triangle BAC is equal to tlie triangle EDG in

all respects : i. 4.

so that the base BC is equal to the base EG.

* See note on the next page.

42 EUCLID'S ELEMENTS.

Again, in the triangle FDG,

because DG is equal to DF,

therefore the angle DFG is equal to the angle DGF, i. 5.

but the angle DGF is greater than the angle EGF;

therefore also the angle DFG is greater than the angle EGF;

still more then is the angle EFG greater than the angle EGF.

And in the triangle EFG,

because the angle EFG is greater than the angle EGF,

therefore the side EG is greater than the side EF ; i. 19.

but EG was shewn to be equal to BC ;

therefore BC is greater than EF. q.e.d.

* This condition was inserted by Simson to ensure that, in the

complete construction, the point F should fall beloxc EG. Without

this condition it would be necessary to consider three cases: for F

might fall above, or upon, or below EG j and each figure would require

separate proof.

We are however scarcely at liberty to employ Simson' s condition

without proving that it fulfils the object for which it was introduced.

This may be done as follows :

Let EG, DF, produced if necessary, intersect at K.

Then, since DE is not greater than DF,

that is, since DE is not greater than DG,

therefore the angle DGE is not greater than the angle DEG. i. 18.

But the exterior angle DKG is greater than the angle DEK : i. 16.

therefore the angle DKG is greater than the angle DGK.

Hence DG is greater than DK. i. 19.

But DG is equal to DF ;

therefore DF is greater than DK.

So that the point F must fall beloio EG.

BOOK I. PROP. 24.

Or the following method may be adopted.

Proposition 24. [Alternative Proof.]

In the triangles ABC, DEF,

let BA be equal to ED,

and AC equal to DF,

but let the angle BAC be greater than

the angle EDF:

then shall the base BC be greater than

the base EF.

For apply the triangle DEF to the

triangle ABC, so that D may fall on A,

and DE along AB:

then because DE is equal to AB,

therefore E must fall on B.

And because the angle EDF is less than the angle BAC,

therefore DF must fall between AB and AC.

Let DF occupy the position AG.

Case I. If G falls on BC :

Then G must be between B and C;

therefore BC is greater than BG.

But BG is equal to EF :

therefore BC is greater than EF.

Case II. If G does not fall on BC.

Bisect the angle CAG by the straight line AK

which meets BC in K.

Join G K.

Then in the triangles GAK, CAK,

/ GA is equal to CA,

Because \ ^"^ ^K is cominon to both;

land the angle GAK is equal

\ angle CAK;

therefore GK is equal to CK.

But in the triangle BKG,

the two sides BK, KG are together greater than the third side BG, i. 20.

that is, BK, KC are together greater than BG ;

therefore BC is greater than BG, or EF. q.e.d.

44 euclid's elements.

Proposition 25. Theorem.

If two triangles have two sides of the one equal to two

sides of the other^ each to each, hut the base of one greater

than tlie base of the other ; then the angle contained by the

sides of that which has t/ie greater ba^e, shall be greater than

the angle contained by the corresponding sides of the other.

Let ABC, DEF be two triangles which have the two sides

BA, AC equal to the two sides ED, DF, each to each, but the

base BC greater than the base EF :

then shall the angle BAC be greater than the angle EDF.

Proof For if the angle BAC be not greater than the

angle EDF, it must be either equal to, or less than the

angle EDF.

But the angle BAC is not equal to the angle EDF,

for then the base BC would be equal to the base EF ; i. 4.

but it is not. J^yp-

Neither is the angle BAC less than the angle EDF,

for then the base BC would be less than the base EF ; i. 24.

but it is not. Ilyp-

Therefore the angle BAC is neither equal to, nor less than

the angle EDF ;

that is, the angle BAC is greater than the angle EDF. q.e.d.

exercise.

In a triangle ABC, the vertex A is joined to X, the middle

point of the base BC; shew that the angle AXB is obtuse or acute,

EXERCISES ON PROPOSITIONS 1 TO 12.

1. Shew tbat the straight line which joins the vertex of an

isosceles triangle to the middle point of the base is perpendicular

to the base.

2. Shew that the straight lines which join the extremities of the

base of an isosceles triangle to the middle points of the opposite sides,

are equal to one another.

* 3. Two given points in the base of an isosceles triangle are equi-

distant from the extremities of the base : shew that they are also equi-

distant from the vertex.

4. If the opposite sides of a quadrilateral are equal, shew that the

opposite angles are also equal.

" 5. Any two isosceles triangles XAB, YAB stand on the same base

AB: shew that the angle XAY is equal to the angle XBY; and that

the angle AXY is equal to the angle BXY.

6. Shew that the opposite angles of a rhombus are bisected by the

diagonal which joins them.

' 7. Shew that the straight lines which bisect the base angles of an

isosceles triangle form with the base a triangle which is also isosceles.

8. ABC is an isosceles triangle having AB equal to AC ; and the

angles at B and C are bisected by straight lines which meet at O :

shew that OA bisects the angle BAC.

/ 9. â– Shew that the triangle formed by joining the middle points of

the sides of an equilateral triangle is also equilateral.

I 10. The equal sides BA, CA of an isosceles triangle BAC are pro-

duced beyond the vertex A to the points E and F, so that AE is equal

to AF; and FB, EC are joined: shew that FB is equal to EC.

11. Shew that the diagonals of a rhombus bisect one another at

right angles.

12. In the equal sides AB, AC of an isosceles triangle ABC two

points X and Y are taken, so that AX is equal to AY; and CX and BY

are drawn intersecting in O : shew that

(i) the triangle BOC is isosceles;

(ii) AC bisects the vertical angle BAC ;

(iii) AG, if produced, bisects BC at right angles.

13. Describe an isosceles triangle, having given the base and the

length of the perpendicular drawn from the vertex to the base.

14. In a given straight line find a point that is equidistant from

two given points.

In what case is this impossible ?

EUCLID'S ELEMENTS.

Proposition 13. Theorem.

If one straight line stand upon anotlier straight line,

tlien the adjacent angles shall he either tivo right angles, or

together equal to two right angles.

A E A

Let the straight line AB stand upon the straight line DC :

then the adjacent angles DBA, ABC shall be either two right

angles, or together equal to two right angles.

Case I. For if the angle DBA is equal to tlie angle ABC,

each of them is a right angle. Def. 7.

Case II. But if the angle DBA is not equal to the

angle ABC,

from B draw BE at right angles to CD. i. 11.

Proof. Now the angle DBA is made up of the two

angles DBE, EBA;

to each ofÂ» these equals add the angle ABC;

then the two angles DBA, ABC are together equal to the

three angles DBE, EBA, ABC. Ax. 2.

Again, the angle EBC is made up of the two angles EBA,

ABC;

to each of these equals add the angle DBE.

Then the two angles DBE, EBC are together equal to the

three angles DBE, EBA, ABC. Ax. 2.

But the two angles DBA, ABC have been shewn to be equal

to the same three angles ;

therefore tlie angles DBA, ABC are together equal to the

angles DBE, EBC. Ax. 1.

But the angles DBE, EBC are two right angles; Constr.

therefore the angles DBA, ABC are together equal to two

light angles. Q. e. d.

BOOK I. PROr. 13.

Definitions.

(i) The complement of an acute angle is its defect from

a right angle, that is, the angle by which it falls short of a right

angle.

Thus two angles are complementary, when their sum is a

right angle.

(ii) The supplement of an angle is its defect from two right

angles, that is, the angle by which it falls short of two right

angles.

Thus two angles are supplementary, when their sum is two

right angles.

Corollary. Arigles which are comide^nentary or supple-

mentary to the same angle are equal to one another.

EXERCISES.

1. If the two exterior angles formed by producing a side of a tri-

angle both ways are equal, shew that the triangle is isosceles.

2. The bisectors of the adjacent angles which one straight line

makes with another contain a right angle.

Note. Jn the adjoining figure AOB

is a given angle; and one of its arms AO

is produced to C: the adjacent angles

AOB, BOC are bisected by OX, OY.

Then OX and OY are called respect-

ively the internal and external bisectors

of the angle AOB.

Hence Exercise 2 may be thus enunciated :

The internal and external bisectors of an angle are at right angles

to one another.

3. Shew that the angles AOX and COY are complementary.

4. Shew that the angles BOX and COX are supplementary; and

also that the angles AOY and BOY are supplementary.

^ ' Euclid's elemei^ts.

v^J*ROPOSiTiON 14. Theorem.

Jj\ at a point hi <i straight Ihie, ttvo other stra'ujht lineSy

on opjiosite sides of it, make the adjacent angles togetlter

equal to ttvo right angles, then these two straight lines shall

he in one and the same straight liv.e.

At the point B in the straight line AB, let the two

straight lines BC, BD, on the opposite sides of AB, make

the adjacent angles ABC, ABD together equal to two right

angles :

then BD shall be in the same straight line with BC.

Proof. For if BD be not in the same straight line with BC,

if possible, let BE be in the same straight line with BC.

Then because AB meets the straight line CBE,

therefore the adjacent angles CBA, ABE are together equal

to two right angles. i. 13.

But the angles CBA, ABD are also together equal to two

right angles. Hyj).

Therefore the angles CBA, ABE are together equal to the

angles CBA, ABD. Ax. 11.

From each of these equals take the common angle CBA;

then the remaining angle ABE is equal to the remaining angle

ABD; the part equal to the whole; wliich is impossible.

Therefore BE is not in the same straight line with BC.

And in the same way it may be shewn that no other

line but BD can be in the same straight line with BC.

Therefore BD is in the same straight line with BC. q.e.d,

EXERCISE. y^

A BCD is a rhombus; and the diagonal AC is bisected at O. If O

is joined to the angular points B and D; shew that OB and CD are

in one straight line.

BOOK I. PROP. 15. .31

Ohs. When two straight lines intersect at a point, four

angles are formed; and any two of these angles which are not

adjacent, are said to be vertically opposite to one another.

Proposition 15. Theorem.

If two straight lines intersect dne'ctHother, theyi the verticallij

vpi^osite angles shall he equal.

A

Let the two straight lines AB, CD cut one another at

the point E :

then shall the angle A EC be equal to the angle DEB,

and the angle CEB to the angle A ED.

Proof. Because AE makes with CD the adjacent angles

CEA, AED,

therefore these angles are together equal to two right

angles. 1. 13.

Again, because DE makes with AB the adjacent angles AED,

DEB,

therefore these also are together equal to two right angles.

Therefore the angles CEA, AED are together equal to the

angles AED, DEB.

From each of these equals take the common angle AED;

then the remaining angle CEA is equal to the remaining

angle DEB. Ax. 3.

In a similar way it may be shewn that the angle CEB

is equal to the angle AED. q.e.d.

Corollary 1. From this it i^ manifest that, if two

straight liyies cut one another, tJie angles which they make

at the point where they cut, are together equal to four right

angles.

Corollary 2. Consequently, when ayiy number of straight

lines meet at a point, the sum of the angles made by con-

secutive Ihies is equal to four right angles.

. EUCLID'S ELEMENTS.

Proposition 16. Theorem.

If one aide of a triangle he j}roduced, then the exterior a^ujle

slmll he greater than either of the interior opposite amjles.

Because

Let ABC be a triangle, and let one side BC be produced

to D : then shall the exterior angle ACD be greater than

either of the interior opposite angles CBA, BAG.

Construction. Bisect AC at E : i. 10.

Join BE; and produce it to F, making EF equal to BE. l 3.

Join FC.

Proof. Then in the triangles AEB, CEF,

AE is equal to CE, Constr.

and EB to EF ; Constr.

also the angle AEB is equal to tlie vertically

opposite angle CEF; I. 15.

therefore the triangle AEB is equal to the triangle CEF in

all respects : i. 4.

so that the angle BAE is equal to the angle ECF.

But the angle ECD is greater than its part, the angle ECF;

therefore the angle ECD is greater than the angle BAE;

that is, the angle ACD is greater than the angle BAC.

In a similar way, if BC be bisected, and the side AC

produced to G, it may be shewn that the angle BCG is

greater than the angle ABC.

But the angle BCG is equal to the angle ACD: i. 15.

therefore also the angle ACD is greater than the angle ABC.

Q. E. D.

book i. prop. 17. 33

Proposition 17. Theorem.

Any two angles of a triangle are together less than two

right angles.

A k

B

Let ABC be a triangle; then shall any two of its angles, as

ABC, ACB, be together less than two riljht angles.

Construction. Produce the side BC to D.

Proof. Then because ACD is an exterior angle of the

triangle ABC,

therefore it is gieater than the interior opposite angle

ABC. I. 16.

To each of these add the angle ACB :

then the angles ACD, ACB are together greater than the

angles ABC, ACB. Ax. 4.

But the adjacent angles ACD, ACB are together equal to

two right angles. i. 13.

Therefore the angles ABC, ACB are together less than two

right angles.

Similarly it may be shewn that the angles BAC, ACB, as

also the angles CAB, ABC, are together less than two right

angles. q. e. d.

Note. It follows from this Proposition that every triangle must

have at least two acute angles : for if one angle is obtuse, or a right

angle, each of the other angles must be less than a right angle.

EXERCISES.

1. Enunciate this Proposition so as to shew that it is the converse l^

of Axiom 12.

2. If any side of a triangle is produced both ways, the exterior

angles so formed are together greater than two right angles. *

3. Shew how a proof of Proposition 17 may be obtained by

joining each vertex in turn to any point in the opposite side.

H. E. *)

34 EUCLID'S ELEMENTS.

Proposition 18. Theorem.

If one side of a triangle be greater than another, then

the angle opposite to the greater side shall be greater than

the angle opposite to the less.

Let ABC be a triangle, in wliich the side AC is greater

than the side AB :

then shall the angle ABC be greater than the angle ACB.

Construction. From AC, the greater, cut off a part AD equal

to AB. L 3.

Join BD,

Proof. Then in the triangle ABD,

because AB is equal to AD,

therefore the angle ABD is equal to the angle ADB. i. 5,

But the exterior angle ADB of the triangle BDC is

greater than the interior opposite angle DCB, that is,

greater than the angle ACB. i. 16.

TJierefore also the angle ABD is greater than the angle ACB;

still more then is the angle ABC greater than the angle

ACB. ^ Q.E.D.

. Euclid enunciated Proposition 18 as follows:

The greater side of every tricmgle has the greater angle

opposite to it.

[This form of enunciation is found to be a common source of diffi-

culty with beginners, who fail to distinguish what is assumed in it and

what is to be proved.]

[For Exercises see page 38.]

I

BOOK I. PROP. 19. 35

Proposition 19. Theorem.

If one angle of a triangle he greater than another^ then

the side opjjosite to the greater angle shall he greater than

the side opposite to the less.

I

Let ABC be a triangle in wliich the angle ABC is greater

tlian the angle ACB ;

then shall the side AC be greater than the side AB.

Proof For if AC be not greater than AB,

it must be either equal to, or less than AB.

But AC is not equal to AB,

for then the angle ABC would be equal to the angle ACB j i. 5.

but it is not. U^Vl^-

Neither is AC less than AB ;

for then the angle ABC would be less than the angle ACB ; I.18.

but it is not : HyP-

Therefore AC is neither equal to, nor less than AB.

That is, AC is greater than AB. q.e.d.

Note, The mode of demonstration used in this ProiDosition is

known as the Proof by Exhaustion. It is applicable to cases in which

one of certain mutually exclusive suppositions must necessarily be

true; and it consists in shewing the falsity of each of these supposi-

tions in turn loith one exception: hence the truth of the remaining

supposition is inferred.

Euclid enunciated Proposition 1 9 as follows :

The greater angle of every triangle is suhtended hy the

greater side, or, has the greater side opposite to it.

[For Exercises see page 38.]

3â€”2

36 Euclid's elements.

Proposition 20. Theorem.

Any two sides of a tt'iangle are together greater than the

third side.

B

Let ABC l)e a triangle: â€¢

then sliall any two of its sides be together greater tlian the

third side :

name]}', BA, AC, shall be greater than CB ;

AC, CB greater than BA ;

and CB, BA greater than AC.

Construction. Produce BA to the point D, making AD equal

to AC. 1. 3.

Join DC.

Proof. Then in the triangle ADC,

because AD is equal to AC, Constr.

therefore the angle ACD is equal to the angle ADC. i. 5.

But the angle BCD is greater than the angle ACD ; Ax. 9.

therefore also the angle BCD is greater than the angle ADC,

that is, than the angle BDC.

And in the triangle BCD,

because the angle BCD is greater than the angle BDC, Pr.

therefore the side BD is greater than the side CB. i. 19.

But BA and AC are together equal to BD ;

therefore BA and AC are together greater than CB.

Similarly it may be shewn

that AC, CB are together greater than BA ;

and CB, BA are together greater than AC. Q. E. D.

[For Exercises aee page 38.]

jioOK I. PROP. i^l. 37

Propositiox 21. Theorem.

If from the ends of a side of a triangle, there he drawn

two straight lines to a j^oint withiri the triangle, then these

straight lines shall be less than the other two sides of the

triangle, hut shall contain a greater angle.

I

Let ABC be a. triangle, aiid from B, C, the ends of tlie

. side BC, let the two straight lines BD, CD be drawn to

a point D wdthin the triangle :

then (i) BD and DC shall be together less than BA and AC ;

ii) the angle BDC shall be greater than the angle BAC.

onstruction. Pi-oduce BD to meet AC in E.

Proof (i) In the triangle BAE, the two sides BA, AE an^

together greater than the third side BE : i, 20.

to each of these add EC :

then BA, AC are together greater than BE, EC. Ax. 4.

Again, in the triangle DEC, the two sides DE, EC are to-

gether greater than DC : - i. 20.

to each of these add BD ;

then BE, EC are together greater than BD, DC.

But it has been shewn that BA, AC are together greater

than BE, EC :

still more tlien are BA, AC greater than BD, DC.

ii) Again, the exterior angle BDC of the triangle DEC is

greater than the interior opposite angle DEC ; i. 16.

and the exterior angle DEC of the triangle BAE is greater

than the interior opposite angle BAE, that is, than the

angle BAC ; i. 16.

still more then is the angle BDC greater than the angle BAC.

Q.E.D.

L th

%

38 Euclid's elements,

exercises

oil PUOPOSITIONS 18 AND 19.

1. The hypotenuse is the greatest side of a right-angled triangle.

2. If two angles of a triangle are equal to one another, the sides

also, which subtend the equal angles, are equal to one another. Prop. 6.

Prove this indirectly by using the result of Prop. 18.

3. BC, the base of an isosceles triangle ABC, is produced to any

point D ; shew that AD is greater than either of the equal sides.

4. If in a quadrilateral the greatest and least sides are opposite to

one another, then each of the angles adjacent to the least side is

greater than its opposite angle.

'â– 5. 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.

â– 6. ABC is a triangle, in which OB, GO bisect the angles ABC.

ACB respectively: shew that, if AB is greater than AC, then OB is

greater than OC.

OK Proposition 20.

!

7. The difference of any two sides of a triangle is less than

the third side.

8. In a quadrilateral, if two opposite sides which are not parallel

are produced to meet one another; shew that the perimeter of the

greater of the two triangles so fonned is greater than the perimeter of

the quadrilateral.

^. The sum of the distances of any point from the three angular

points of a triangle is greater than half its perimeter.

10. The perimeter of a quadrilateral is greater than the sum of its

diagonals.

11. Obtain a proof of Proposition 20 by bisecting an angle by a

straight line which meets the opposite side.

ox Proposition 21.

12. In Proposition 21 shew that the angle BDC is greater than

the angle BAC by joining AD, and producing it towards the base.

13. The sum of the distances of any point within a triangle from

its angular points is less than the perimeter of the triangle.

BOOK I. PROP. 22.

39

Proposition 22. Problem.

To describe a triangle having its sides equal to three

given straight lines, any two of which are together greater

than the third.

B

Let A, B, C be the three given straight lines, of which

any two are together greater than the third.

It is required to describe a triangle of which the sides

shall be equal to A, B, C.

Construction. Take a straight line DE terminated at tlie

point D, but unlimited towards E.

Make DF equal to A, FG equal to B, and GH equal to C. I. 3.

From centre F, with radius FD, describe the circle DLK.

From centre G with radius GH, describe the circle MHK,

cutting the former circle at K.

Join FK, GK.

Then shall the triangle KFG have its sides equal to the

tliree straight lines A, B, C.

Proof. Because F is the centre of the circle DLK,

therefore FK is equal to FD : Def 11.

but FD is equal to A ; Constr.

. therefore also FK is equal to A. Ax. 1.

Again, because G is the centre of the circle MHK,

therefore GK is equal to GH : Def. 11.

but G H is equal to C ; Constr.

therefore also GK is equal to C. Ax. 1.

And FG is equal to B. Constr.

Therefore the triangle KFG has its sides KF, FG, GK equal

respectively to the three given lines A, B, C. q.e.f.

40 EUCLIDS KLEilKNTS.

Â£XERCISÂ£.

Ou a given base describe a triangle, whose remaining sides shall be

equal to two given straight lines. Point out how the construction

fails, if any one of the three given lines is greater than the sum of

the other two.

Proposition 23. Problem.

At a given point in a given straigJtt line, to make an

angle equal to a given angle.

Let AB be the given straight line, and A the given point

in it; and let DCE be the given angle.

It is required to draw from A a straight line making

with AB an angle equal to the given angle DCE.

Construction. In CD, CE take any points D and E ;

and join DE.

From AB cut off AF equal to CD. I. 3.

On AF describe the triangle FAG, having the remaining

sides AG, GF equal respectively to CE, ED. i. 22.

Then shall the angle FAG be equal to the angle DCE.

Proof. For in the triangles FAG, DCE,

( FA is equal to DC, Constr.

Because < and AG is equal to CE; Constr.

(and the base FG is equal to the base DE : Constr.

therefore the angle FAG is equal to the angle DCE. i. 8.

That is, AG makes with AB, at the given point A, an angle

equal to the given angle DCE. q.e.f.

BOOK I. PROP. 24. 41

PROPOSITION 24.

Jf two triangles have two sides of the one equal to two

sides of the other, each to each, but the angle contained by

the two sides of one gi-eater than the angle contained by

the corresponding sides of the other ; then the base of that

which has the greater angle shall be greater than the base of the

other.

D'

Let ABC, DEF be two triangles, in wliicli the two sides

BA, AC are equal to the two sides ED, DF, each to each,

but the angle BAC greater than the angle EDF :

then shall the base BC be greater than the base EF.

' * Of the two sides DE, DF, let DE be that which is not

greater than DF.

Construction. At the point D, in the straight line ED,

and on the same side of it as DF, make the angle EDG

equal to the angle BAC. ]. 23.

Make DO equal to DF or AC; i. 3.

and join EG, GF.

Proof Then in the triangles BAC, EDG,

I' BA is equal to ED, HyjJ.

-D and AC is equal to DG, Constr.

I also the contained angle BAC is equal to the

\ contained angle EDG ; Constr.

Therefore the triangle BAC is equal to tlie triangle EDG in

all respects : i. 4.

so that the base BC is equal to the base EG.

* See note on the next page.

42 EUCLID'S ELEMENTS.

Again, in the triangle FDG,

because DG is equal to DF,

therefore the angle DFG is equal to the angle DGF, i. 5.

but the angle DGF is greater than the angle EGF;

therefore also the angle DFG is greater than the angle EGF;

still more then is the angle EFG greater than the angle EGF.

And in the triangle EFG,

because the angle EFG is greater than the angle EGF,

therefore the side EG is greater than the side EF ; i. 19.

but EG was shewn to be equal to BC ;

therefore BC is greater than EF. q.e.d.

* This condition was inserted by Simson to ensure that, in the

complete construction, the point F should fall beloxc EG. Without

this condition it would be necessary to consider three cases: for F

might fall above, or upon, or below EG j and each figure would require

separate proof.

We are however scarcely at liberty to employ Simson' s condition

without proving that it fulfils the object for which it was introduced.

This may be done as follows :

Let EG, DF, produced if necessary, intersect at K.

Then, since DE is not greater than DF,

that is, since DE is not greater than DG,

therefore the angle DGE is not greater than the angle DEG. i. 18.

But the exterior angle DKG is greater than the angle DEK : i. 16.

therefore the angle DKG is greater than the angle DGK.

Hence DG is greater than DK. i. 19.

But DG is equal to DF ;

therefore DF is greater than DK.

So that the point F must fall beloio EG.

BOOK I. PROP. 24.

Or the following method may be adopted.

Proposition 24. [Alternative Proof.]

In the triangles ABC, DEF,

let BA be equal to ED,

and AC equal to DF,

but let the angle BAC be greater than

the angle EDF:

then shall the base BC be greater than

the base EF.

For apply the triangle DEF to the

triangle ABC, so that D may fall on A,

and DE along AB:

then because DE is equal to AB,

therefore E must fall on B.

And because the angle EDF is less than the angle BAC,

therefore DF must fall between AB and AC.

Let DF occupy the position AG.

Case I. If G falls on BC :

Then G must be between B and C;

therefore BC is greater than BG.

But BG is equal to EF :

therefore BC is greater than EF.

Case II. If G does not fall on BC.

Bisect the angle CAG by the straight line AK

which meets BC in K.

Join G K.

Then in the triangles GAK, CAK,

/ GA is equal to CA,

Because \ ^"^ ^K is cominon to both;

land the angle GAK is equal

\ angle CAK;

therefore GK is equal to CK.

But in the triangle BKG,

the two sides BK, KG are together greater than the third side BG, i. 20.

that is, BK, KC are together greater than BG ;

therefore BC is greater than BG, or EF. q.e.d.

44 euclid's elements.

Proposition 25. Theorem.

If two triangles have two sides of the one equal to two

sides of the other^ each to each, hut the base of one greater

than tlie base of the other ; then the angle contained by the

sides of that which has t/ie greater ba^e, shall be greater than

the angle contained by the corresponding sides of the other.

Let ABC, DEF be two triangles which have the two sides

BA, AC equal to the two sides ED, DF, each to each, but the

base BC greater than the base EF :

then shall the angle BAC be greater than the angle EDF.

Proof For if the angle BAC be not greater than the

angle EDF, it must be either equal to, or less than the

angle EDF.

But the angle BAC is not equal to the angle EDF,

for then the base BC would be equal to the base EF ; i. 4.

but it is not. J^yp-

Neither is the angle BAC less than the angle EDF,

for then the base BC would be less than the base EF ; i. 24.

but it is not. Ilyp-

Therefore the angle BAC is neither equal to, nor less than

the angle EDF ;

that is, the angle BAC is greater than the angle EDF. q.e.d.

exercise.

In a triangle ABC, the vertex A is joined to X, the middle

point of the base BC; shew that the angle AXB is obtuse or acute,

Online Library → Euclid → A text-book of Euclid's Elements : for the use of schools : Books I-VI and XI → online text (page 3 of 27)