### Classical Geometry

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6.5
MISCELLANEOUS TOPICS
Morley's Theorem
The following result was discovered by Frank Morley in about 1900. He mentioned
it to friends in Cambridge and published it about 20 years later in Japan.
Morley's Theorem states that the points of intersection of the adjacent trisectors of
the angles of any triangle are the vertices of an equilateral triangle, as in the figure
below.
A
Before proving this theorem we need a lemma, which is yet another characterization
of the incenter of a triangle.
Lemma 6.5.1. (Another Characterization of the lncenter)
The incenter of a triangle 6ABC is the unique point I interior to the triangle which
satisfies the following two properties:
(I) it lies on an angle bisector of one of the angles (say at A) and
(2) it subtends an angle 90 + ~LA with the side BC.
Proof. The incenter has property ( 1) by definition, since it is the intersection of the
internal angle bisectors of the triangle.
MORLEY'S THEOREM
179
A
B
c
D
Also, from the External Angle Theorem, we have
LBIC
= LBID + LCID
=x+y+x+z
=x+
1
2(2x+2y+2z)
1
=X+ 2180
=
90+x
1
= 90+ 2LA,
and property (2) holds also.
To prove uniqueness, suppose that the point I' "f. I lies on the angle bisector of LA,
and suppose that I' also subtends an angle 90 + ~LA with the side BC.
A
B
D
c
180
MISCELLANEOUS TOPICS
Note that in the figure on the previous page, where I' is between A and I, the External
Angle Inequality implies that
s
> s'
and
t > t',
so that
90+
1
1
2LA = LBIC = s +t > s' +t' = LBI'C = 90+ 2LA,
which is a contradiction. Similarly, if I is between A and I', we again get a
contradiction. Therefore, the point I satisfying (1) and (2) is unique.
0
Theorem 6.5.2. (Morley's Theorem)
The points of intersection of adjacent trisectors of the angle of any triangle form an
equilateral triangle.
Proof. Let 6ABC be a fixed triangle, so that the angles at the vertices A, B, and
C are fixed angles. It is enough to prove Morley's Theorem for a triangle 6A' B' C'
that is similar to the given triangle, since scaling the sides by a proportionality factor
k does not change the angles, and so an equilateral triangle remains equilateral.
In the proof we start with an equilateral triangle 6PQ R and then construct a triangle
6A' B' C' that is similar to 6ABC and has the property that the adjacent trisectors
form the given equilateral triangle 6PQ R.
Construction Phase
Step 1. Construct an equilateral triangle 6PQ R.
p
60
60
60
R
Q
MORLEY'S THEOREM
181
Step 2. Construct three isosceles triangles on the sides of 6.PQR, with angles as
shown,
R'
R
where
1
3
'
1
60- -LB
3
'
o:
= 60- -LA
(3
=
1
'Y=60--LC
3
.
Observations:
(1) 0
< o:, (3, 'Y < 60.
Since, for example, LA
LA< 180 implies that o:
(2)
0:
> 0 implies that 60 > 60 - ~LA = o:,
= 60- ~LA > 60- ~ 180 = 60- 60 = 0.
+ (3 + 'Y = 120.
Sinceo:+fJ+'Y
= 180- HLA+LB+LC) = 180- 1 ~ 0 = 180-60 = 120.
(3) The sum of any two of the angles o:, (3, 'Y is greater than 60.
Since, for example,
o: + (3
while
= 120- ~(LA+ LB) >
120- 1 ~ 0
= 120-60 = 60.
182
MISCELLANEOUS TOPICS
Step 3. Extend the sides of the isosceles triangles until they meet as shown to produce
a larger triangle DA' B' C'.
A'
We claim that DA' B' C' is similar to DABC and that the lines
A'P,
A'Q,
B'P,
B'R,
C'Q,
C'R
are the angle trisectors of the angles at A', B', and C'.
Argument Phase
Step 1. a
+ (3 + 1 + 60 =
180, so that the angles are as shown on the figure.
For example, the angle LB' P R' is a straight angle and L_Q P R = 60 so that a +
60 + (3 + LB' PQ' = 180, which implies that 1 = LB' PQ' and that the vertically
opposite angle LA' P R' = 1 also.
We should also show that the extensions of the sides of the isosceles triangles actually
do intersect at the points A', B', and C'.
For example, if we isolate part of the figure, we can show that Q R' and Q' R intersect
at a point C', as shown in the figure on the following page.
MORLEY'S THEOREM
If we consider the sum L.RQR'
L.RQR'
183
+ L.Q'RQ, then
+ L.Q' RQ =a+ 60 + ,6 + 60 =a+ ,6 + 120 > 60 + 120 = 180,
and therefore the parallel postulate says that RQ' and R' Q intersect on the side of
the transversal Q R where the sum of the interior angles is less than 180; that is, RQ'
and R' Q intersect on the side opposite P at some point C', as shown.
R'
R
Similarly, PQ' and P' Q intersect at some point A', while P R' and P' R intersect at
some point B', as shown.
Step 2.
a+ ,6 + 'Y = 120, and therefore
L.PA'Q
= 60- a= ~L.A,
L.P B' R
= 60 - ,6 = ~ L.B,
L.QC' R
= 60 -
'Y
= ~ L.C.
For example, in L:.P B' R, the sum of the interior angles is
L.P B' R
+ 'Y + ,6 + ,6 + a = 180,
so that
L.PB' R
= 180-,6- (a+ ,6 + 'Y) = 180-,6- 120 = 60- ,6,
and
L.P B' R = 60 - ,6 =
~ L.B.
Similarly,
L.P A' Q
= 60 - a = ~ L.A and L.QC' R = 60 -
'Y
= ~ L.C.
184
MISCELLANEOUS TOPICS
Step 3. R is the incenter of D.B' R' C'. Similarly, P is the incenter of D. A' B' P',
while Q is the incenter of D.A' C' Q'.
We will show that R is the incenter of D.B' R' C'. The other two results follow in the
same way.
(a) Note that R lies on the angle bisector of LB' R'C' since D.P R' R is congruent
to D.Q R' R by the SSS congruency theorem.
R'
B'
C'
(b) Note that
LB'RC' = 180-a = 90+(90-a) =
90+~(180-2a) = 90+~LB'R'C'.
By the characterization theorem for the incenter proven in the lemma, (l) and (2)
imply that R is the incenter of D.B' R' C'.
Therefore,
LPB'R= LRB'C' = !LB
3
so that P B' and RB' are angle trisectors of LB'. Similarly, P A' and Q A' are angle
trisectors of LA', and QC' and RC' are angle trisectors of LC'.
Therefore, LA' = LA, LB' = LB, and LC' = LC, and by the AAA similarity
theorem, D. A' B' C' is similar to D. ABC, and the corresponding segments in D. ABC
are the angle trisectors. Therefore, in D.ABC the points of intersection of adjacent
trisectors of the angles form an equilateral triangle.
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