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CIHM/ICMH
Microfiche
Series.
CIHM/ICMH
Collection de
microfiches.
Canadian Institute for Historical Microreproductions
Institut Canadian de microreproductions historiques
1980
Technical Notes / Notes techniques
The Institute has attempted to obtain the best
original copy available for filming. Physical
features of this copy which may alter any of the
images in the reproduction are checked below.
m
D
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Coloured covers/
Couvertures de couleur
Coloured maps/
Cartes gdographiques en couleur
Pages discoloured, stained or foxed/
Pages d6colordes. tachetdes ou piqu6es
Tight binding (may cause shadows or
distortion along interior margin)/
Reliure serrd (peut causer de I'ombre ou
de la distortion le long de la marge
int^rieure)
L'Institut a microfilm^ le meilleur exemplaire
qu'il lui a 6t6 possible de se procurer. Certains
ddfauts susceptibles de nuire d la quality de la
reproduction sont notds ci-dessous.
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D
m
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Coloured pages/
Pages de couleur
Coloured plates/
Planches en couleur
Show through/
Transparence
Pages damaged/
Pages endommagdes
The
pos
oft
film
The
con
or t
app
The
filrr
insl
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in (
upf
bot
foil
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Additional comments/
Commentaires suppldmentaires
Bibliographic Notes / Notes bibliographiques
n
Only edition available/
Seule Edition disponible
D
Pagination incorrect/
Erreurs de pagination
□
Bound with other material/
Relid avec d'autres documents
Cover title missing/
Le titre de couverture manque
n
Pages missing/
Des pages manquent
Maps missing/
Des cartes gdographiques manquent
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Plates missing/
Des planches manquent
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Additional comments/
Commentaires suppldmentaires
The images appearing here are the best quality
possible considering the condition and legibility
of the original copy and in keeping with the
filming contract specifications.
Les images suivantes ont 6t6 reproduites avec le
plus grand soin, compte tenu de la condition et
de la nettet6 de I'exemplaire film6, et en
conformity avec les conditions du contrat de
filmage.
The last recorded frame on each microfiche shall
contain the symbol —»► (meaning CONTINUED"),
or the symbol V (meaning "END"), whichever
applies.
Un des symboles suivants apparaTtra sur la der-
nidre Image de cheque microfiche, selon le cas:
le symbols — ► signifie "A SUIVRE", le symbole
y signifie "FIN".
The original copy was borrowed from, and
filmed with, the kind consent of the following
institution:
National Library of Canada
L'exemplaire film6 fut reproduit grfice d la
g6n6rosit6 de I'dtabiissement prdteur
suivant :
Bibliothdque nationale du Canada
Maps or plates too large to be entirely included
in one exposure are filmed beginning in the
upper left hand corner, left to right and top to
bottom, as many frames as required. The
following diagrams illustrate the method:
Les cartes ou les planches trop grandes pour dtre
reproduites en un seul clich6 sont filmdes d
partir de I'angle sup6rieure gauche, de gauche d
droite et de haut en bas, en prenant le nombre
d'images n6cessaire. Le diagramme suivant
illustre la mdthode :
1
2
3
1 2 3
4 5 6
EXTRAORDINARY
TRICAL DISCOVERY
TRISEGTION
OF ANY RECTILINEAL ANGLE
— BY
ELEMENTARY GEOMETRY
«
AND SOLUTIONS OF OTHER PROBLEMS
Considered impomble except by aid of the higher Geometry,
— BY —
AjYDREW' DOYLE.
'--'-'^'^S^ ^?2/»t-V-t
OTTAWA:
A. BUREAU, PRINTEB. SPARKS STREET
1880
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aJ ,.^>-<jd.'''£^^ P'<^-<>\_-<j^'<L-l^
4.A^r
«» ^ «**»««»«•»♦*• <»«*» a « • «*|»-5
EXTRAORDINARY
GEOMETRICAL DISCOVERY
TRISECTION
OF ANY RECTILINEAL ANGLE
— BY —
ELEMENTARY GEOMETEY
AND SOLUTIONS OF OTHER PROBLEMS
Considered impossible except by aid of the higher Geometry.
— BY
ANDREW DOYLE.
C-».'Vl^g^:^t/^U..-J
OTTAWA :
A. BUREAU, PRINTER, SPARKS STREET.
1880
"
)
31 ^/>
Eiitoi'cd ftccovding to Act of Parliament of Canada, in the year
1880, ^y AndrbW Doylb, in the office of the Miniator of Agriculture.
\
Trisection of a Rectilineal Angle.
Let B O C be the aiiujlo to be triwected.
At the point O, make (23.1) the aii.^les BOD and C
O J*^, each equal to the angle B O C. Take O as centre
with any radius OB, and describe the arc DBCE;
join D B, B 0, C E, and D K. Produce I) B, and E C to
meet in T; join O T. By 4.1., the angle O D B is
equal to the angle O K C, and (5.1) the angle O D E is
equal to the angle O E D ; take away the two latter
angles and the remaining angles E D T and I) E T are
equal : therefore (0.1) I) T is equal to E T. In the
triangles O D T and O E T, the three sides of the one
are respectively equal to the three sides of the other,
by (8.1) they are equal in every respect; therefore the
angle D T O is equal to the angle E T O, then the line
O T i)isecting the vertical angle of an isosceles triangle,
bisects D E and B C perpendicularly. It also bisects
the arc BO; it is evident by 4.1 that D B, B C, and C
E are equal to one another.
Draw a d parallel to B C. so that a d shall be equal a
B and d C. This is done by bisecting the angles T B C
and T C B by lines meeting the sides B T and C T in
a and d, when the triangle is isosceles, or equilateral
and proved hy (29.1, 6.1, and 4.1). Find the diamater
of a circle circumscribing the trapizium B a (Z C, and
make t L equal to it ; join a L and d L, cutting the arc
B C in the points r and I : these are the trisecting points
of the arc B C which is t measure of the given angle
B O C. Draw the lines Orb ard O I c, and the three
angles 6 c, 6 O B, and C O c are equal and each
equal to one third of the given angle B C.
.
H E
FIRST DEMONSTEATION.
Through the point B, di'aw B G paralled to O / or O
c, and meeting 1) E in G ; draw also B P parallel to O r
or O 6; make (3.1) B A equal to B G ; through G
draw G H parallel to B O, and through the point A
draw A H parallel to B G ; join B H, and B H passew
through the point P. We have now, the straight line
B P and the line B H, having the two points B and P
in common ; therefore (Prop. 2, Legendre,) they coin
cide throughout the whole, or they are in one and the
same straight line.
A B is parallel to G H, and A H is parallel to B G ;
therefore A B G H is a parallelogram, having its op-
{DObite sides equal and parallel, and A B is equal to B G
)y construction ; therefore A B G H is a rhombus and
the diagonal B H bisects its opposite angles ; therefore
the angle A B P is equal to the angle GBP.
B G is parallel to O / and B O a line meeting them,
by (29.1), the angle A B G is equal to the angle BO/;
for the same reason, the angle A B P is equal to the
angle B O r ; therefore the remaining angle G B P is
equal to the angle r O I but GBP and A B P are equal ;
B or and roe similarly it can be proved that the angle I O
C is equal to the angle r O I] but things which are equal
to the same thing, are equal to one another; therefore
the angles B O r, r O /, and I O C are equal, and the
angle B O C is trisected.
SECOND SEMONSTBATION.
Suppose L C and L B to be joined, it is easily proved
that L C is equal to L B, then by 5.1, the angle L B C
is equal to the angle L C B, and the angle B C T is
equal to the angle C B T ; therefore the whole angle
6
Jj C d in equal to the whole angle L B a. In the
triangles d C Jj and a Blj, a B and B L are equal to d
and C L, and their contained angles equal, by (4.1),
a L Ih equal to d L, and the angle B a L equal to the
angle C d L, by (13.1), the angle r a b 'i» equal to the
angle / d c. In the triangle r L O and I L O, the three
sides of the one are respectively equal to the three sides
of the other, by (8.1), O r L is equal to the angle O / L,
and the angle r O L eqiuil to the angle / O L; by (15.1)
the angle orb and die are equa but it has been
proved that the angle barm equal lo c d I; therefore
(32.1), the two triangles are equiangular, and a r equal
\o d i\ therefore, by (2().l), a 6 is equal to c d^ but a B
is equal d C ; therefore b B is equal to c C. Then, in
the triangles 6 B O and c C O, we have b B and B O
equal to c C and C O and their contained angles equal,
by (4.1), the angle B O 6 is equal to the angle C O c.
Then, as in the first demonstration it may be proved
that each of these angles^ is equal to the angle 6 0^;
therefore the three angles B O /, r O /, and I O C,
are equal, and the angle B O C is trisected.
THIED DEMONSTRATION.
Join B I and C r, cutting O / and O r in n and m. If
on O B we describe a semicircle, it passes through the
point m; therefore (31.3), the angles at m are right
angles, and (3.3) B I is bisected perpendicularly by O
m ; hence B m and m O are equal to O m and m I and
the angles at m equal, by (4.1), the angle B O m is
equal to the angle I O m. This may also be proved by
letting fall a perpendicular from O, on B Z and this per-
pendicular always falls on the point m ; then we have
two straight lines which coinside in part they must
coinside throughout the whole (Legendre, Prop. 2). It
may be similarly proved that r is bisected perpendi-
cularly by O c in w, and that the angle C O n is equal
to the angle r O w ; therefore the three angles, B O r,
r O /, and / O C, are equal ; therefore the angle B O C
is trisected.
POUETH DEMONSTEATION.
In O T, take any point, as L, and with L O as radius
describe the circle O U V S. Through Y draw V X
parallel to S O, meeting the circle in X ; and through u
draw u X parallel to V O.
Because S O is parallel to V X and V O meeting
them, (29.1; the angle S O V is equal to the angle O V
X ; for the same reason, the angle O V X is ^^qual to the
angle V X U ; but V X U is equal to V O U ; therefore
the angle S O V is equal to the angle V O U. Similarly
the angle C O U may be proved equal to V O U ; there-
fore S O V=V O U=U O C, and the angle B O C is
trisected. The lines V X and U X, in the trisection of
ever}^ angle, always meet the circumference of the circle
in the same point. This evidently, can never happen
but when the three arcs B r, v I, and I C are equal.
ANDREW DOYLE.
w
8
The following problems also have been considered by
all mathematicians to be beyond the range of elemen-
tary geometry.
1. Let BAG (Fig. 2) be any angle ; take any point
D in A B and let fall the perpendicular D E ; through
D, draw D P parallel to AC; it is required to draw
the line A N O P, so that N P may be equal to twicQ
AD.
Trisect the angle B A C by A P and N P is the line
required. Bisect N P in O, and join O D.
red to
^iv, 80 that
the angle A CI H may be equal to twice the angle
AHG.
Trisect B A C by A P ; take any point G and draw
G II parallel to A P, meeting A C in II.
AN DEE W DOYLE.
d
2. Let C A G be any angle, (Fig. 2) it is requii
raw a line G H cutting the legs of the angle, sc
ll
In the next Edition will appear the method of finding two
geometrican means between two straight lines, and also the dupli-
cation of the cube.
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