Euclid.

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

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Font size Take two-thirds of each of the given medians, and construct
the triangle OKC. The rest of the construction follows easily.]

Vlir. ON AREAS.

See Propositions 35 â€” 48.

It must be understood that throughout this section the word
equal as applied to rectilineal figures will be used as denoting
equality of area unless otherwise stated.

1. Shew that a parallelogram is bisected by any straight line
which passes through the middle point of one of its diagonals, [i. 29,
26.]

2. Bisect a parallelogram by a straight line drawn through a
given point.

3. Bisect a parallelogram by a straight line drawn perpendicular
to one of its sides.

4. Bisect a parallelogram by a straight line drawn parallel to a
given straight line.

5. ABCD is a trapezium in which the side AB is parallel to DC.
Shew that its area is equal to the area of a parallelogram formed by
draioing through X, the middle point of BC, a straight line parallel to

6. A trapezium is equal to a parallelogram whose base is half the
sum of the parallel sides of the given figure, and whose altitude is
equal to the perpendicular distance between them.

7. ABCD is a trapezium in which the side AB is parallel to DC;
shew that it is double of the triangle formed by joining the extremities
of AD to X, the middle point of BC.

8. Shew that a trapezium is bisected by the straight line which
joins the middle points of its parallel sides. [i. 38.]

110 EUCLID'S ELEMENTS,

In the following group of Exercises the proofs depend chiefly
on Propositions 37 and 38, and the two converse theorems.

9. If two straight lines AB, CD intersect at X, and if the straight
lines AC and BD, which join their extremities are parallel, shew that
the triangle AXD is equal to the triangle BXC.

10. If two straight lines AB, CD intersect at X, so that the
triangle AXD is equal to the triangle XCB, then AC and BD are
parallel.

11. ABCD is a parallelogram, and X any point in the diagonal
AC produced; shew that the triangles XBC, XDC are equal. [See
Ex. 13, p. 64.]

12. ABC is a triangle, and R, Q the middle points of the sides
AB, AC; shew that if BQ and CR intersect in X, the triangle BXC is
equal to the quadrilateral AQXR. [See Ex. 5, p. 73.]

13. If the middle points of the sides of a quadrilateral be joined
in order, the parallelogram so formed [see Ex. 9, p. 97] is equal to
half the given figure.

14. Two triangles of equal area stand on the same base but on
opposite sides of it : shew that the straight line joining their vertices
is bisected by the base, or by the base produced.

15. The straight line which joins the middle points of the dia-
gonals of a trapezium is parallel to each of the two parallel sides.

16. (i) A triangle is equal to the sum or difference of two triangle.^
on the same base (or on equal bases), if the altitude of the former is equal
to the sum or difference of the altitudes of the latter.

(ii) A triangle is equal to the sum or difference of two triangles of
the same altitude if the base of the former is equal to tlie sum or differ-
ence of the bases of the latter.

Similar statements hold good of parallelograms.

17. ABCD is a parallelogram, and O is any point outside it;
shew that the sum or difference of the triangles CAB, OCD is equal to
half the parallelogram. Distinguish between the two cases.

On the following proposition depends an important theorem
in I!klechanics : we give a ]:>roof of the first case, leaving the second
case to be deduced by a similar method.

THEOREMS AND EXAMPLES ON BOOK I. Ill

18. (i) ABCD is a parallelogram, and O is any point without the
angle BAD and its opposite vertical angle ; shetc that the triangle OAC
is equal to the sum of the triangles OAD, OAB.

(ii) If O is within the angle BAD or its opposite vertical angle,
the triangle OAC is equal to the difference of the triangles OAD,
OAB.

Cask I. If O is without the z DAB
and its opp. vert, z , then OA is with-
out the par" ABCD : therefore the perp.
drawn from C to OA is equal to the sum
of the perp" drawn from B and D to OA.
[SeeEx. 20, p. 99.]

Now the AÂ« OAC, OAD, OAB are
upon the same base OA ;
and the altitude of the a OAC with
respect to this base has been shewn to
be equal to the sum of the altitudes of
Therefore the A OAC is equal to the sum of the a*Â» OAD, OABÂ».^
[See Ex. 16, p. 110.] q.e.d.

19. ABCD is a parallelogram, and through O, any point within
it, straight lines are drawn parallel to the sides of the parallelogram;
shew that the difference of the parallelograms DO, BO is double of
the triangle AOC. [See preceding theorem (ii).]

20. The area of a quadrilateral is equal to the area of a triangle
having two of its sides equal to the diagonals of the given figure, and
the included angle equal to either of the angles between the dia-
gonals.

21. ABC is a triangle, and D is any point in AB: ii is required to
draw through D a straight line DE fo meet BO produced in E, so that
the triangle DBE may he equal to the triangle ABC.

[Join DC. Through A draw AE parallel to DC. i. 31.

Join DE.
The A EBD shall be equal to the a ABC. i. 37.]

11*2 Euclid's elements.

22. On a base of given length describe a triangle equal to a given
triangle and having an angle equal to an angle of the given triangle.

23. Construct a triangle equal in area to a given triangle, and
having a given altitude.

24. On a base of given length construct a triangle equal to a
given triangle, and having its vertex on a given straight line.

25. On a base of given length describe (i) an isosceles triangle ;
(ii) a right-angled triangle, equal to a given triangle.

26. Construct a triangle equal to the sum or difference of two
given triangles. [See Ex. 1(5, p. IIO.J

27. ABC is a given triangle, and X a given point: describe a
triangle equal to ABC, having its vertex at X, and its base in the samo
straight line as BC.

28. A BCD is a quadrilateral: on the base AB construct a triangle
equal in area to A BCD, and having the angle at A common with the

[Join BD. Through C draw CX parallel to BD, meeting AD pro-
duced in X ; join BX, ]

29. Construct a rectilineal figure equal to a given rectilineal
figure, and having fetoer sides by one than the given figure.

Hence shew how to constnict a triangle equal to a given rectilineal
figure.

30. A BCD is a quadrilateral : it is required to construct a triangle
eciual in area to ABCD, having its vertex at a given point X in DC,
and its base in the same straight line as AB.

31. Construct a rhombus equal to a given parallelogram.

32. Construct a parallelogram which shall have the same area
and perimeter as a given triangle.

33. Bisect a triangle by a straight line draicn through one of itn
angular points.

34. Trisect a triangle by straight lines drawn through one of its
angular points. [See Ex. 19, p. 102, and i. 38.]

35. Divide a triangle into any number of equal parts by straight
lines drawn through one of its angular points.

[See Ex. 19, p. 99, and i. 38.]

THEOREMS AND EXAMPLES ON BOOK I.

113

3G. Bisect a triangle bij a straight, line draivn through a given
point in one of its sides.

[Let ABC be the given a , and P the
given point in the side AB.

Bisect AB at Z ; and join CZ, CP.

Through Z draw ZQ parallel to CP.

Join PQ,

Then shall PQ bisect the a .

See Ex. 21, p. 111.]

37. Trisect a triangle by straight lines drawn from a given point ir
one of its sides.

[Let ABC be the given a , and X the given
point in the side BC.

Trisect BC at the points P, Q. Ex. 19, p. 99.
Join AX, and through P and Q draw PH
and QK parallel to AX.

Join XH, XK.
These straight lines shall trisect the a ; as
may be shewn by joining AP, AQ.

See Ex. 21, p. 111.]

P X Q C

38. Cut off from a given triangle a fourth, fifth, sixth, or any
part required by a straight line drawn from a given point in one of its
sides. [See Ex. 19, p. 99, and Ex. 21, p. 111.]

39. Bisect a quadrilateral by a straight line drawn through an
angular point.

[Two constructions may be given for this problem : the first will
be suggested by Exercises 28 and 33, p. 112.

The second method proceeds thus.

Let A BCD be the given quadrilateral,
and A the given angular point.

Join AC, BD, and bisect BD in X.
Through X draw PXQ parallel to AC,

meeting BC in P; join AP.
Then shall AP bisect the quadrilateral.
Join AX, CX, and use i. 37, 38.]

40. Cut off from a given quadrilateral a third, a fourth, a fifth, or
any part required, by a straight line drawn through a given angular
point. [See Exercises 28 and 35, p. 112.]

H. E.

8

114 Euclid's elements.

[The following Theorems depenil on i. 47.]

41. In the figure of i. 47, shew that

(i) the sum of the squares on AB and AE is equal to the sum

of the squares on AC and AD,
(ii) the square on EK is equal to the square on AB with four

times the square on AC.
(iii) the sum of the squares on EK and FD is equal to five
times the square on BC.

42. If a straight line he divided into any two parts the square on
the straight line is greater than the squares on the two parts.

43. If the square on one side of a triangle is less than the squares
on the remaining sides, the angle contained by these sides is acute; it'
greater, obtuse.

44. ABC is a triangle, right-angled at A; the sides AB, AC am
intersected by a straight line PQ, and BQ, PC are joined : shew that
the sum of the squares on BQ, PC is equal to the sum of the squares
on BC, PQ.

45. In a right-angled triangle four times the sum of the squares
on the medians which bisect the sides containing the right angle
is equal to five times the square on the hypotenuse.

46. Describe a square whose area shall be three times that of
a given square.

47. Divide a straight line into two parts such that the sum of
their squares shall be equal to a given square.

IX. ON LOCI.

It is frequently required in the course of Plane Geometry to
find the position of a point which satisfies given conditions.
Now all problems of this type hitherto considered have been
found to be capable of definite determination, though some admit
of more than one solution: this however will not be the case if
(yaly one condition is given. For example, if w'e are asked to find
a point which shall be at a given distance from a given point,
we observe at once that the problem is indeterminate, that is,
that it admits of an indefinite number of solutions ; for the
condition stated is satisfied by any point on the circumference
of the circle described from the given point as centre, with a
radius equal to the given distance : moreover this condition is
satisfied by no other point within or without the circle.

Again, suppose that it is required to find a point at a given
distance from a given straight line.

THEOREMS AND EXAMPLES ON BOOK T.

115

Here, too, it is obvious that there are an infinite number of
such points, and that they he on the two parahel straight hues
which may be drawn on either side of the given straight Hne at
the given distance from it : further, no point that is not on one
or other of these parallels satisfies the given condition.

Hence we see that when one condition is assigned it is not
sufficient to determine the position of a point absolutely, but
it may have the effect of restricting it to some definite line or
lines, straight or curved. This leads us to the following definition.

Definition. The Locus of a point satisfying an assigned
condition consists of the line, lines, or part of a line, to which
the point is thereby restricted; provided that the condition is
.satisfied by every point on such line or lines, and by no other.

A locus is sometimes defined as the path traced out by a
point which moves in accordance with an assigned law.

Thus the locus of a point, which is always at a given distance
from a given point, is a circle of which the given point is the
centre : and the locus of a point, which is always at a given distance
from a given straight line, is a pair of parallel straight lines.

We now see that in order to infer that a certain line, or
system of lines, is the locus of a point under a given condition,
it is necessary to prove

(i) that any point which fulfils the given condition is on the
supposed locus ;

(ii) that every point on the supposed locus satisfies the given
condition.

1. Find the locus of a point ivhich is always equidistant from
tioo given points.

Let A, B be the two given points,
(a) Let P be any point equidistant from A
and B, so that AP=BP.

Bisect AB at X, and join PX.
Then in the a^ AXP, BXP,
( AX = BX, Const)'.

Because and PX is common to both,

( alsoAP=BP, Hyp.

.'. the / PXA = the z PXB; i! 8.

and they are adjacent z ^ ;
.-. PX is perp. to AB.
.'. Any point which is equidistant from A and B
is on the straight line which bisects AB at right angles.

116 EUCLID'S ELEMENTS.

(/3) Also every point in this line is equidistant from A and B.

For let Q. be any point in this line.

Join AQ, BQ.

Then in the aÂ« AXQ, BXQ,

r AX = BX,

Because J and XQ is common to both ;

[also the / AXQ = the Z BXQ, being rt. l"*;

.-. AQ=BQ. ].4.

That is, Q is equidistant from A and B.

Hence we conclude that the locus of the point equidistant from
two given points A, B is the straight line which bisects AB at right
angles.

2. To find the locus of the middle point of a straight line drawn
from a giv^^oint to meet a given straight line of unlimited length.

Let A be the given point, and BC the given straight line of un-
limited length.

(a) Let AX be any straight line drawn through A to meet BC,
and let P be its middle point.

Draw AF perp. to BC, and bisect AF at E.

Join EP, and produce it indefinitely.

Since AFX is a a , and E, P the middle points of the two sides AF, AX,

.-. EP is parallel to the remaining side FX. Ex. 2, p. 90.

.-. P is on the straight line which passes through the fi:(ed point E,

and is parallel to BC.

{^) Again, every point in EP, or EP produced, fulfils the required
condition.

For, in this straight line take any point Q.
Join AQ, and produce it to meet BC in Y.
Then FAY is a a , and through E, the middle point of the side AF, EQ
is drawn parallel to the side FY,

. â€˘. Q is the middle point of AY. Ex. 1, p. {

Hence the required locus is the straight line drawn parallel to BC,
and passing through E, the middle point of the perp. from A to BC.

THEOREMS AND EXAMPLES ON BOOK I. 117

3. Find the locus of a point equidistant from two given inter-
secting straight lines. [See Ex. 3, p, 49.]

4. Find the locus of a point at a given radial distance from the
circumference of a given circle.

5. Find the locus of a point which moves so that the sum of its
distances from two given intersecting straight lines of unlimited
length is constant.

6. Find the locus of a point when the differences of its distances
from two given intersecting straight lines of unlimited length is
constant.

7. A straight rod of given length slides between two straight
rulers placed at right angles to one another: find the locus of its
middle point. [See Ex. 2, p. 100.]

8. On a given base as hypotenuse right-angled triangles are
described : find the locus of their vertices.

9. AB is a given straight line, and AX is the perpendicular drawn
from A to any straight line passing through B: find the locus of
the middle point of AX.

10. Find the locus of the vertex of a triangle, when the base and
area are given.

11. Find the locus of the intersection of the diagonals of a paral-
lelogram, of which the base and area are given.

12. Find the locus of the intersection of the medians of a triangle
described on a given base and of given area.

X. ON THE INTERSECTION OP LOCI.

It appears from various problems which have already been
considered, that we are often required to find a point, the
position of which is subject to two given conditions. The method
of loci is very useful in the solution of problems of this kind :
for corresponding to each condition there will be a locus on
which the required point must lie ; hence all points which are
common to these two loci, that is, all the points of intersection
of the loci, will satisfy both the given conditions.

lib EUCLID tt ELEMiiiiMb.

Example 1. To cdhstruct a triangle, having given tfie base, the
altitude, and the length of the median lohich bisects the base.

Let AB be the given base, and P and Q the lengths of the altitude
and median respectively:

then the triangle is known if its vertex is known,
(i) Draw a straight line CD imrallel to AB, and at a distance
from it equal to P :

then tlic required vertex must lie on CD.

(ii). Again, from the middle point of AB as centre, with radius
equal to Q, describe a circle :

then tJie required vertex must lie on this circle.

Hence any points which are common to CD and the circle,
satisfy both the given conditions: that is to say, if CD intersect the
circle in E, F each of the points of intersection might be the vertex
of the required triangle. This supposes the length of the median
Q to be greater than the altitude.

Example 2. To find a point equidistant from three given points
A, B, C, ichich are not in the same straight line.

(i) The locus of points equidistant from A and B is the straight
line PQ, which bisects AB at right angles. Ex. 1, p. 115.

(ii) Similarly the locus of points equidistant from B and C is
the straight line RS w^hich bisects BC at right angles.

Hence the point common to PQ and RS must satisfy both con-
ditions: that is to say, the point of intersection of PQ and RS will
be equidistant from A, B, and C.

These principles may also be used to prove the theorems
relating to concurrency already given on i^age 103.

Example. To prove that the bisectors of^fhe angles of a triangle
are concurrent. \

Let ABC be a triangle.
Bisect the zÂ« ABC, BCA by straight
lines BO, CO: these must meet at
some point O. Ax. 12.

Join OA.
Then shall OA bisect the z BAC.
Now BO is the locus of points equi-
distant from BC, BA; Ex. 3, p. 49.
.-. OP = OR.
Similarly CO is the locus of points
equidistant from BC, CA.

.-. OP = OQ; hence ORr^OQ.
.-. O is on the locus of points equidistant from AB and AC
that is OA is the bisector of the Z BAC.
Hence the bisectors of the three z ^^ meet at the point O.

THEOREMS AND EXAMPLES ON BOOK I. 119

It may happen that the data of the problem are so related
to one another that the resulting loci do not intersect : in this
case the problem is impossible.

For example, if in Ex. 1, j)age 118, the length of the given
median is less than the given altitude, the straight line CD will
not be intersected by the circle, and no triangle can fulfil the
conditions of the problem. If the length of the median is equal
to the given altitude, 07ie point is common to the two loci ;
imd consequently only one solution of the problem exists :
and we have seen that there are two solutions, if the median
is greater than the altitude.

In examples of this kind the student should make a point
of investigating the relations which must exist among the data,
in order that the problem may be possible ; and he must observe
that if under certain relations tivo solutions are possible, and
under other relations no solution exists, there will always be
some intermediate relation under which one and 07il^ one solution
is possible.

EXAMPLES.

1. Find a point in a given straight line which is equidistant
from two given points.

2. Find a point which is at given distances from each of two
given straight lines. How many solutions are possible?

3. On a given base construct a triangle, having given one angle at
the base and the length of the opposite side. Examine the relations
xchich must exist among the data in order that there may be two solu-
tions, one solution, or that the problem may be impossible.

4. On the base of a given triangle construct a second triangle
equal in area to the first, and having its vertex in a given straight
line.

5. Construct an isosceles triangle equal in area to a given
triangle, and standing on the same base.

6. Find a point which is at a given distance from a given point,
and is equidistant from two given parallel straight lines.

BOOK II.

Book II. deals witli the areas of rectangles and squares.

Definitions.

1. A Rectangle is a parallelogram whicli has one of
its angles a right angle.

It should be remembered that if a parallelogram has one right
angle, all its angles are right angles. [Ex. 1, p. 64.]

2. A rectangle is said to be contained by any two of
its sides which form a right angle : for it is clear that both
the form and magnitude of a rectangle are fully determined
when the lengths of two such sides are given.

Thus the rectangle ACDB is said
to be contained by AB, AC; or by CD,
DB : and if X and Y are two straight
lines equal respectively to AB and AC,
then the rectangle contained by X and Y
is equal to the rectangle contained by
AB, AC.

[See Ex. 12, p. 64.]

x-

Y-

After Proposition 3, we sliall use the abbreviation
reel. AB, AC to denote tlie rectangle contained hi/ ^B and
AC.

3. In any parallelogram the figure formed by either
of the parallelograms about a diagonal together with the
two complements is called a gnomon.

Thus the shaded portion of the annexed
figure, consisting of the parallelogram EH
together with the complements AK, KC is
the (fnomon AHF.

The other gnomon in the figure is that
which is made up of AK, GF and FH,
namely the gnomon AFH,

introductory. 121

Introductory.

Pure Geometry makes no use of number to estimate the
magnitude of the lines, angles, and figures with which it deals :
hence it requires no imits of magnitude such as the student is
familiar with in Arithmetic.

For example, though Geometry is concerned with the relative
lengths of straight lines, it does not seek to express those lengths
in terms of yards^ feet, or inches: similarly it does not ask how
many square yards or square feet a given figure contains, nor how
many degrees there are in a given angle.

This constitutes an essential difierence between the method
of Pure Geometry and that of Arithmetic and Algebra ; at the
same time a close connection exists between the results of these
two methods.

In the case of Euclid's Book II., this connection rests ujjon
the fact that the number of units of area in a rectangidar figure
is found by multiplying together the numbers of units of length in

For example, if the two sides AB, AD ,
of the rectangle A BCD are respectively
four and three inches long, and if through
the points of division parallels are drawn
ns in the annexed figure, it is seen that
the rectangle is divided into three rows,
each containing four square inches, or
into four columns, each containing three
square inches. " "

Hence the whole rectangle contains 3x4, or 12, square
inches.

Similarly if AB and AD contain m and n units of length
respectively, it follows that the rectangle A BCD will contain nin
units of area: further, if AB and AD are equal, each containing
m units of length, the rectangle becomes a square, and contains
m^ units of area.

[It must be understood that this explanation implies that the
lengths of the straight lines AB, AD are commensurable, that is, that
they can be expressed exactly in terms of some common unit.

This however is not always the case: for example, it may be
proved that the side and diagonal of a square are so related, that
it is impossible to divide either of them into equal parts, ofiohich the
other contains an exact number. Such lines are said to be incommen-

122 Euclid's elements.

surable. Hence if the adjacent sides of a rectangle are incommen-
surable, we cannot choose any linear unit in terms of which these
sides maybe exactly expressed; and thus it will be impossible to sub-
divide the rectangle into squares of unit area, as illustrated in the
figure of the preceding page. We do not here propose to enter
further into the subject of incommensurable quantities : it is suffi-
cient to point out that further knowledge of them will convince the
student that the area of a rectangle may be expressed to any required
degree of accuracy by the product of the lengths of two adjacent
sides, whether those lengths are commensurable or not. ]

From the foregoing explanation we conclude that the rectangle
attained by two straight lines in Geometry corresponds to the
product of two numbers in Arithmetic or Algebra ; and that the
square described on a straight line corresponds to the square of

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