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Euclid.

Euclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools online

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Online LibraryEuclidEuclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools → online text (page 18 of 38)
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by the lines joining their centers, has its angles equal to those in the
segments.

110. The perpendiculars let fall from the three angles of any tri-
angle upon the opposite sides, intersect each other in the same point.

111. If AD J CE be drawn perpendicular to the sides BC, AB of



ON BOOK 111. '^mmm j-yj

le triangle ABC, prove that the rectangle contained by JBC and^D,
is equal to the rectangle contained by £A and JBJE.

112. The lines which bisect the vertical angles of all triangles on the
same base and -with the same vertical angle, all intersect in one point.

113. Of all triangles on the same base and between the same
parallels, the isosceles has the greatest vertical angle.

114. It is required within an isosceles triangle to find a point such,
that its distance from one of the equal angles may be double its dis-
tance from the vertical angle. ^

115. To find within an acute-angled triangle, a point from which,
if straight lines be drawn to the three angles of the triangle, they shall
make equal angles with each other.

116. A flag-staif of a given height is erected on a tower whose
height is also given : at what point on the horizon will the flag-staft
appear under the greatest possible angle ?

117. A ladder is gradually raised against a wall ; find the locus of
its middle point.

118. The triangle formed by the chord of a circle (produced
or not), the tangent at its extremity, and any line perpendicular
to the diameter through its other extremity, will be isosceles.

119. AD, JBJE are perpendiculars from the angles A and JB
on the opposite sides of a triangle, BF perpendicular to JEJD or JED
produced ; shew that the angle FBD = EBA,

XI.

120. If three equal circles have a common point of intersection
])rove that a straight line joining any two of the points of intersectioi.^
will be perpendicular to the straight line joining the other two points
of intersection.

121. Two equal circles cut one another, and a third circle touches
each of these two equal circles externally ; the straight line which joins
the points of section will, if produced, pass through the center of the
third circle.

122. A number of circles touch each other at the same point, and a
straight line is drawn from it cutting them : the straight lines joining
each point of intersection with the center of the circle will be all parallel.

123. If three circles intersect one another, two and two, the three
chords joining the points of intersection shall all pass through one
point.

124. If three circles touch each other externally, and the three
common tangents be drawn, these tangents shall intersect in a point
equidistant from the points of contact of the circles.

125. If two equal circles intersect one another in A and B, and
from one of the points of intersection as a center, a circle be described
which shall cut both of the equal circles, then will the other point of
intersection, and the two points in which the third circle cuts the
other two on the same side of AB, be in the same straight line.

XIL

126. Given the base, the vertical angle, and the difference of the
sides, to construct the triangle.

t2



172 GEOMETRICAL EXERCISES

127. Describe a triangle, having given the vertical angle, and
the segments of the base made by a line bisecting the vertical angle.

128. Given the perpendicular height, the vertical angle and the
sum of the sides, to construct the triangle.

129. Construct a triangle in which the vertical angle and the
difference of the two angles at the base shall be respectively equal to
two given angles, and whose base shall be equal to a given straight
line.

130. Given the vertical angle, the difference of the two sides con-
taining it, and the clifference of the segments of the base made by a
perpendicular from the vertex ; construct the triangle.

131. Given the vertical angle, and the lengths of two lines drawn
from the extremities of the base to the points of bisection of the sides,
to construct the triangle.

132. Given the base, and vertical angle, to find the triangle whose
area is a maximum.

133. Given the base, the altitude, and the sum of the two re-
maining sides ; construct the triangle.

134. Describe a triangle of given base, area, and vertical angle.

135. Given the base and vertical angle of a triangle, find the
locus of the intersection of perjDendiculars to the sides from the ex-
tremities of the base.

XIII.

136. Shew that the perpendiculars to the sides of a quadrilateral
inscribed in a circle from their middle points intersect in a fixed point.

137. The lines bisecting any angle of a quadrilateral figure in-
scribed in a circle, and the opposite exterior angle, meet in the cir-
cumference of the circle.

138. If two opposite sides of a quadrilateral figure inscribed in a
circle be equal, prove that the other two are parallel,

139. The angles subtended at the center of a circle by any two
opposite sides of a quadrilateral figure circumscribed about it, are
together equal to two right angles.

140. Four circles are described so that each may touch internally
three of the sides of a quadrilateral figure, or one side and the ad-
jacent sides produced ; shew that the centers of these four circles will
all lie in the circumference of a circle.

141. One side of a trapezium capable of being inscribed in a given
circle is given, the sum of the remaining three sides is given ; and also
one of the angles opposite to the given side : construct it.

142. If the sides of a quadrilateral figure inscribed in a circle be
produced to meet, and from each of the points of intersection a
straight line be diawn, touching the cii'cle, the squares of these tan-
gents are together equal to the square of the straight line joining the
points of intersection.

143. If a quadrilateral figure be described about a circle, the
sums of the opposite sides are equal ; and each sum equal to half the
perimeter of the figure.

144. A quadi-ilateral A BCD is inscribed in a circle, BC and DC



ON BOOK III. 173

are produced to meet AD and AB produced in JE and F. The angles
^J5Cand ADC are together equal to AFC, AEB, and twice the
angle ^^C.

145. If the hypotenuse AB of a right-angled triangle ABC he
bisected in Z), and ^Di^ drawn perpendicular to AB, and DE, DF
cut off each equal to DA, and CE, CFjoined, prove that the last two
lines will bisect the angle at C and its supplement respectively.

146. ABCD is a quadrilateral figure inscribed in a circle.
Through its angular points tangents are drawn so as to form another
quadrilateral figure FBLCHDEA circumscribed about the circle.
Find the relation which exists between the angles of the exterior and
the angles of the interior figure.

147. The angle contained by the tangents drawn at the extremi-
ties of any chord in a circle is equal to the difference of the angles in
segments made by the chord : and also equal to twice the angle con-
tained by the same chord and a diameter di-awn from either of its
extremities.

148. li ABCD be a quadrilateral figure, and the lines AB, AC,
AD be equal, shew that the angle BAD is double of CBD and CDB
together.

149. Shew that the four lines which bisect the interior angles of
a quadrilateral figure, form by their intersections, a quadrilateral figure
which can be inscribed in a circle.

150. In a quadrilateral figure ABCD is inscribed a second
quadrilateral by joining the middle points of its adjacent sides ; a
third is similarly inscribed in the second, and so on. Shew that each
of the series of quadrilaterals will be capable of being inscribed in a
circle if the first three are so. Shew also that two at least of the
opposite sides of ABCD must be equal, and that the two squares upon
these sides are together equal to the sum of the squares upon the
other two.

XIV.

151. If from any point in the diameter of a semicircle, there be
drawn two straight lines to the circumference, one to the bisection of
the circumference, the other at right angles to the diameter, the
squares upon these two lines are together double of the square upon
the semi-diameter.

152. If from any point in the diameter of a circle, straight lines
be drawn to the extremities of a parallel chord, the squares on these
lines are together equal to the squares on the segments into which the
diameter is divided.

153. From a given point without a circle, at a distance from the
circumference of the circle not greater than its diameter, draw a
straight line to the concave circumference which shall be bisected by
the convex circumference.

154. If any two chords be drawn in a circle perpendicular to
each other, the sum of their squares is equal to twice the square on
the diameter diminished by four times the square on the line joining
the center with their point of intersection.



174 GEOMETRICAL EXERCISES ON BOOK 111.

155. Two points are taken in the diameter of a circle at any
equal distances from the center ; through one of these draw any chord,
and join its extremities and the other point. The triangle so formed
has the sum of the squares of its sides invariable.

156. If chords drawn from any jfixed point in the circumference
of a circle, be cut by another chord which is parallel to the tangent
at that point, the rectangle contained by each chord, and the part of
it intercepted between the given point and the given chord, is constant.

157. If AB be a chord of a circle inclined by half a right angle to
the tangent at A, and A C, AD be any two chords equally inclined to
AB,ACUAD' = 2.AB\

158. A chord POQ cuts the diameter of a circle in Q, in an angle
equal to half a right angle ; P0^+0Q^ = 2 (rad.)^

159. Let ACDB be a semicircle whose diameter is -45; and
AD, i?C any two chords intersecting in P; prove that

AB'= - DA.AP+CB.BP,

160. If ABDChe any parallelogram, and if a circle be described
passing through the point A, and cutting the sides AB, AC, and the
diagonal AD, in the points F, G, H respectively, shew that

AB.AF^^AC.AG^-AD.AH.

1 61. Produce a given straight line, so that the rectangle under the
given line, and the whole line produced, may equal the square of the
part produced.

162. If ^ be a point within a circle, J?Cthc diameter, and through
A, AD be drawn perpendicular to the diameter, and BAE meeting
the circumference in E, then BA.BE=BC.BD.

163. The diameter A CD of a circle, whose center is C, is pro-
duced to P, determine a point F in the line A P such that the rectangle
PF. PC may be equal to the rectangle PD.PA.

164. To produce a given straight line, so that the rectangle con-
tained by the whole line thus produced, and the part of it produced,
shall be equal to a given square.

165. Two straight lines stand at right angles to each other, one of
which passes through the center of a given circle, and from any point
in the other, tangents are drawn to the circle. Prove that the chord
joining the points of contact cuts the first line in the same point, what-
ever be the point in the second from which the tangents are di'awn.

166. A, B, C, D, are four points in order in a straight line, find
a point J5^ between B and C, such that AE.EB = ED.EC, by a
geometrical construction.

167. If any two circles touch each other in the point 0, and lines
be drawn through O at right angles to each other, the one line cutting
the circles in P, P, the other in Q, Q' ; and if the line joining the
centers of the circles cut them in A, A' ; then

PP'^QQ^ = A'A\




BOOK IV.



DEFINITIONS.



A RECTILINEAL figure is said to be inscribed in another rectilineal
figure, when all the angular points of the inscribed figure are upon
the sides of the figure in which it is inscribed, each upon each.




In like manner, a figure is said to be described about another fignire,
when all the sides of the circumscribed figure pass through the angular
]Doints of the figure about which it is described, each through each.

III.

A rectilineal figure is said to be inscribed in a circle, when all the
angular points of the inscribed figure are upon the circumference of
the circle.




A rectilineal figure is said to be described about a circle, when each
side of the circumscribed figure touches the circumference of the cii'cle



In like manner, a circle is said to be inscribed in a rectili«ieal figure,
when the circumference of the circle touches each side of the figure.

VI.

A circle is said to be described about a rectilineal figure, when the
circumference of the circle passes through all the angular points of
the fi":ure about which it is described.



176



VII.



A straight line is said to be placed in a circle, when the extremities
of it are in the circumference of the cuxle.



PROPOSITION I. PROBLEM.

In a given circle to place a straight line, equal to a given straight line
which is not greater than the diameter of the circle.

Let ABC he the given circle, and D the given straight line, not
greater than the diameter of the circle.

It is required to place in the circle ABC a straight line equal to D.

A




Draw BCthe diameter of the circle ABC.

Then, if ^Cis equal to D, the thing required is done;

for in the circle ABC a straight line BC is placed equal to D.

But, if it is not, BCis greater than D ; (hyp.)

make CJE equal to J), (l. 3.)

and from the center C, at the distance CU, describe the circle AEF,

and join CA.

Then CA shall be equal to D.

Because C is the center of the circle AUF,

therefore CA is equal to CJE : (i. def. 15.)

but CU is equal to D ; (constr.)

therefore D is equal to CA. (ax. 1.)

Wherefore in the circle ABC, a straight line CA is placed equal to

the given straight line Z), which is not greater than the diameter of the

circle. Q.E.F.

PROPOSITION II. PROBLEM.

In a given circle to inscribe a triangle equiangular to a given triangle.

Let ABC he the given circle, and DEF the given triangle.
It is required to inscribe in the circle ABC a triangle equiangular
to the triangle DEF.




Draw the straight line GAHioMchmg the circle in the point A, (ill. 17.)
and at the point A, in the straight line AH,



BOOK IV. PROP. 11, III. 177

make the angle HA C equal to the angle DUF-, (i. 23.)
VKk and at the point A, in the straight line A G,

I^B make the angle GAB equal to the angle BFE;

vKk and join JBC-. then ^jBC shall be the triangle required.
Up Because HAG touches the circle ABC,

and ^ C is drawn from the point of contact,
tlierefore the angle HACis equal to the angle ABC in the alternate
segment of the circle : (lii. 32.)

but HACis equal to the angle DBF-, (constr.)

therefore also the angle ABC is equal to DBF: (ax. 1.)

for the same reason, the angle ACB is equal to the angle BFE:

therefore the remaining angle BA C is equal to the remaining angle

FBF\ (I. 32. and ax. 3.)

wherefore the triangle ABC\^ equiangular to the triangle BEF,

and it is inscribed in the circle ABC. Q.E.r.

PROPOSITION III. PROBLEM.
About a given circle to describe a triangle equiangular to a given triangle.

Let ABC he the given circle, and BEF the given triangle.
It is required to describe a triangle about the circle ABC equian-
gular to the triangle BEF.
L




Produce EF both ways to the points G, H\

find the center K of the circle ABC, (iii. 1.)

and from if draw any straight line KB ;

at the point K in the straight line KB,

make the angle BKA equal to the angle BEG, (l. 23.) -

and the angle BKC equal to the angle BFH-,

and through the points A, B, C, draw the straight lines Z-4Jf, ilOiV,

NCL, touching the circle ABC. (ill. 17.)

Then XiUfiV shall be the triangle required.
Because LM, MN, NL touch the circle ABC in the points A, B,
C, to which from the center are drawn KA, KB, KC,
tlierefore the angles at the points^, B, Care right angles: (ill. 18.)
and because the four angles of the quadrilateral figure AMBK are
equal to four right angles,

for it can be divided into two triangles ;

and that two of them KAM, KBIIare right angles,

therefore the other two AKB, AMB are equal to two right angles :



(ax. 3.)
the ane



but the angles BEG, BEF are likewise equal to two right angles j
(I. 13.)

1 5



178 Euclid's elements.

therefore the angles AKB, AMB are equal to the angles DEG, DEF-,
(ax. 1.)

of which AKB is equal to DEG-, (constr.)
wherefore the remaining angle AMB is equal to the remaining angle
DEF. (ax. 3.)
In like manner, the angle LNM may be demonstrated to be equal
toDi^^;

and therefore the remaining an^rle MLN is equal to the remaining

angle EDF: (l 32 and ax. 3.)
therefore the triangle EMJV is equiangular to the triangle DEF:
and it is described about the circle ABC. Q.E.F.

PROPOSITION IV. PROBLEM.

To inscribe a circle in a given triangle.

Let the given triangle he ABC.
It is required to inscribe a circle in ABC




Bisect the angles ABC, i5Cu4 by the straight lines^Z), CD meeting
one another in the point D, (i. 9.)

from which draw DE, DF, DG perpendiculars to AB, BC,CA. {l. 12.)
And because the angle EBB is equal to the angle FBD,
for the angle ABC is bisected by BD,
and that the right angle BED is equal to the right angle BED-, (ax. 11.)
therefore the two triangles EBD, FBD have two angles of the one
equal to two angles of the other, each to each ;
and the side BD, which is opposite to one of the equal angles in each,
is common to both ;

therefore their other sides are equal ; (l. 26.)
wherefore DE is equal to DF:
for the same reason, DG is equal to DF:
therefore DE is equal to DG: (ax. 1.)
therefore the three straight lines DE, DF, DG are equal to one
another ;
and the circle described from the center D, at the distance of any
of them, will pass through the extremities of the other two, and
touch the straight lines AB, BC, CA,

because the angles at the points E, F, G are right angles,
and the straight line which is drawn from the extremity of a diameter

at right angles to it, touches the circle : (ill. 16.)
therefore the straight lines AB, BC, CA do each of them touch the

circle,
and therefore the circle EFG is inscribed in the triangle ABC. q.e.f.



BOOK IV. PROP. V, VI.



179






PROPOSITION V. PROBLEM.

To describe a circle about a given triangle.

Let the given triangle hQ ABC.
It is required to describe a circle about ABC.

Bl-^ ^*h ^C »

Bisect AB, AC in the points D, E, (i. 10.)

and from these points draw DF, BF at right angles to AB, A C; (l. 11.)

DF, EF produced meet one another :

for, if they do not meet, they are parallel,

wherefore AB, AC, which are at right angles to them, are parallel ;

which is absurd :

let them meet in F, and join FA ;
also, if the point F be not in BC, join BF, CF.
Then, because AD is equal to DB, and EF common, and at right
angles to AB,

therefore the base ^J^is equal to the base FB. (l. 4.)

In like manner, it may be shewn that CF is equal to FA ;

and therefore ^jPis equal to FC; (ax. 1.)

and FA, FB, FC are equal to one another :

wherefore the circle described from the center F, at the distance of

one of them, will pass through the extremities of the other two, and

be described about the triangle ABC Q.e.f.

Cor. — And it is manifest, that when the center of the circle falls
within the triangle, each of its angles is less than a right angle, (ill. 31.)
each of them being in a segment greater than a semicircle ; but, when
the center is in one of the sides of the triangle, the angle opposite to
this side, being in a semicircle, (ill. 31.) is a right angle; and, if the
center falls without the triangle, the angle opposite to the side beyond
which it is, being in a segment less than a semicircle, (ill. 31.) is greater
than a right angle : therefore, conversely, if the given triangle be
acute-angled, the center of the circle falls within it ; if it be a right-
angled triangle, the center is in the side opposite to the right angle ;
and if it be an obtuse-angled triangle, the center falls without the tri-
angle, beyond the side opposite to the obtuse angle.

PROPOSITION VI. PROBLEM.

To inscribe a square in a given circle.

Let ABCD be the given circle.

It is required to inscribe a square in ABCD,

A




180 Euclid's elements.

Draw the diameters, A C, BD, at right angles to one another, (ill. 1.
and L 11.)

and join AB, BC, CD, DA.
The figure A BCD shall be the square required.
Because BE is equal to ED, for E is the center, and that EA is
common, and at right angles to BD ;

the base BA is equal to the base AD : (l. 4.)
and, for the same reason, BC, CD are each of them equal to BA,
or AD ;

therefore the quadrilateral figure ABCD is equilateral.

It is also rectangular ;

for the straight line BD being the diameter of the circle ABCD,

BAD is a semicircle ;

wherefore the angle BAD is a right angle : (ITL 31.)

for the same reason, each of the angles ABC, BCD, CD A is a right

angle :

therefore the quadrilateral figure ABCD is rectangular:

and it has been shewn to be equilateral,

therefore it is a square : (l. def. 30.)

and it is inscribed in the circle ABCD. q.e.F.

PROPOSITION VII. PROBLEM.
To describe a sq>j,are about a given circle.

Let ABCD be the given circle.
It is required to describe a square about it.

G A F



^


-^


V


J



H C K

Draw two diameters AC, BD of the circle ABCD, at right angles
to one another,
and through the points A, B, C, D, draw EG, GH, HK, KF touch-
ing the circle. (lii. 17.)

The figure GIIKF shall be the square required.
Because EG touches the circle ABCD, and EA is drawn from the
center E to the point of contact A,
therefore the angles at A are right angles: (ill. 18.)
for the same reason, the angles at the points B, C, D are right angles ;
and because the angle AEB is a right angle, as likewise is EBG,
therefore GHh parallel to AC: (l. 28.)
for the same reason ^Cis parallel to FK:
and in like manner GF, UK may each of them be demonstrated to
be parallel to BED :
therefore the figures GK, GC, AK, FB, BK^re parallelograms;

and therefore GFi^ equal to HK, and GH to FK-. {I. 34.)
and because ACh equal to BD, and that ACi^ equal to each of the
two GH, FK',



BOOK IV. PROP. VII, VIII. 181

and BD to each of the two GF, II K:

Gil, FKave each of them equal to GF, or UK;

therefore the quadrilateral figure FGIIK is equilateral.

It is also rectangular ;

for GBEA being a parallelogram, and AEB a right angle,

therefore A Gli is likewise a right angle : (l. 34.)

and in the same manner it may be shewn that the angles at H, K, F^

are right angles :

therefore the quadrilateral figure FGHK is rectangular ;

and it was demonstrated to be equilateral ;

therefore it is a square ; (l. def. 30.)

and it is described about the circle A BCD. q.e.f.



PROPOSITION YIII. PROBLEM.

To inscribe a circle in a given square.

Let ABCD be the given square.
It is required to inscribe a circle in ABCD,



Ha

vI3



Bisect each of the sides AB, AD in the points F, F, (l. 10.)
and through F draw ^^ parallel to AB or DC, (i. 3L)
and through i^draw i^X parallel to AD or BC:
therefore each of the figures AK, KB, AH, IID, AG,GC,BG, GD
is a right-angled parallelogram ;

and their opposite sides are equal : (i. 34.)

and because AD is equal to AB, (i. def. 30.)

and that AE is the half of AD, and AF the half of AB,

therefore AE is equal to AF; (ax. 7.)

wherefore the sides opposite to these are equal, viz. EG to GE :

in the same manner it may be demonstrated that GH, GK are each

of them equal to FG or GE :
therefore the four straight lines GE, GF, GH, GK are equal to one

another ;
and the circle described from the center G at the distance of one of
them, will pass through the extremities of the other three, and touch
the straight lines AB, BC, CD, DA ;

because the angles at the points E, F, H, K, are right angles, (l. 29.)

and that the straight line which is drawn from the extremity of a

diameter, at right angles to it, touches the circle : (ill. 16. Cor.)

therefore each of the straight lines AB, BC, CD, DA touches the circle,

which therefore is inscribed in the square ABCD. q.e.f.



182



PROPOSITION IX. PROBLEM.

To describe a circle about a given square.

Let ABCD be the given square.
It is required to describe a circle about ABCD,
A^ D '




Join A C, BD, cutting one another in JE :

and because DA is equal to AB, and -4 C common to the triangles

DAC,BAC, (I. def. 30.)

the two sides DA, AC are equal to the two BA, A C, each to each ;

and the base DC is equal to the base BC;

wherefore the angle DA C is equal to the angle BAC; (l. 8.)

and the angle DAB is bisected by the straight line AC:
in the same manner it may be demonstrated that the angles ABC,
BCD, CD A are severally bisected by the straight lines BD,AC:



Online LibraryEuclidEuclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools → online text (page 18 of 38)