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### Proof of Morley's Theorem

```
Date: 08/09/2000 at 00:31:34
Subject: Trisecting angles

If you take any triangle ABC and trisect each angle,

ABK = KBL = LBC
ACJ = JCL = LCB
BAK = KAJ = JAC

Each pair of trisectors meet a point, forming an equilateral triangle
JKL. Can you prove it? Thanks.

B
/^\
/    \
/    L  \
/ K*  *    \
/     * J     \
A /_ _ _ _ _  _ _ _\C

```

```
Date: 08/09/2000 at 07:54:38
From: Doctor Floor
Subject: Re: Trisecting angles

Thanks for writing. You are referring to the famous "Morley trisector
theorem," by Frank Morley (1904).

Note that I use DEF for your JKL.

We will use the Law of Sines, which states that

a       b       c
----- = ----- = ----- = 2R
sin(A)  sin(B)  sin(C)

and thus, for instance, also

a = 2R sin(A)

where a = BC, b = AC, c = AB, and R is the radius of the circumcircle
of triangle ABC.

We apply this to triangle BDC and find

BD               a                  a
-------- = -------------------- = -------------
sin(C/3)   sin(180 - B/3 - C/3)   sin(120 + A/3)

where we use the fact that A+B+C = 180 degrees.

We rewrite this to

a sin(C/3)     2R sin(A) sin(C/3)
BD = -------------- = ------------------ ..................[1]
sin(120 + A/3)     sin(120 + A/3)

Now we will use the following trigonometric identities:

4 sin(t)sin(t+60)sin(t+120)

= 4 sin(t)(sin(t)cos(60) + cos(t)sin(60))*(sin(t)cos(120)
+ cos(t)sin(120))

= 4 sin(t)(1/2  sin(t) + 1/2 sqrt(3)cos(t))*(-1/2 sin(t)
+ 1/2 sqrt(3)cos(t))

= 4 sin(t)(3/4 cos^2(t) - 1/4 sin^2(t))

= 4 sin(t)(1/2 cos^2(t) + 1/4(cos^2(t) - sin^2(t)))

= 2 sin(t)cos^2(t) + sin(t)cos(2t)

= sin(2t)cos(t) + sin(t)cos(2t)

= sin(3t)

So [1] can be rewritten as

BD = 8R sin(A/3)sin(C/3)sin(60 + A/3)

In the same way we find

BF = 8R sin(A/3)sin(C/3)sin(60 + C/3)

and we can apply the Law of Cosines to triangle BDF:

DF^2 = BD^2 + BF^2 - 2*BD*BF*cos(B/3)

= 64R^2 sin^2(A/3)sin^2(C/3)*(sin^2(60+A/3) +
sin^2(60+C/3) - 2 sin(60+A/3)sin(60+C/3)cos(B/3))  ...[2]

Now again we will have to apply a trigonometric identity, using the
fact that the sum of 60+A/3, 60+C/3 and B/3 is 180 degrees. I will
show below that if u+v+w = 180 degrees, then we find:

sin^2(u) + sin^2(v) - sin^2(w) = 2 sin(u)sin(v)cos(w)   .....[3]

Applying [3] to [2] we find that

DF^2 = 64R^2 sin^2(A/3)sin^2(B/3)sin^2(C/3)

and thus

DF = 8R sin(A/3)sin(B/3)sin(C/3)

Since this expression is independent of permutations of A,B and C, we
will find the same expression for DE and EF, so that indeed DEF is an
equilateral triangle.

********

Proof of identity [3].

I recall from the following message from the Dr. Math archives,
"Equilateral Triangles: Area Formula and Proof,"

http://mathforum.org/dr.math/problems/andy6.16.98.html

that if x+y+z = 2*pi = 360 degrees then

cos^2(x) + cos^2(y) + cos^2(z) = 1 + 2cos(x)cos(y)cos(z)

We apply this to u+90, v+90 and w, and find, using that
cos(t+90)= -sin(t) for any angle t.

sin^2(u) + sin^2(v) + cos^2(w) = 1 + 2 sin(u)sin(v)cos(w)

sin^2(u) + sin^2(v) + cos^2(w) =
sin^2(w) + cos^2(w) + 2 sin(u)sin(v)cos(w)

sin^2(u) + sin^2(v) - sin^2(w)= 2 sin(u)sin(v)cos(w)

which proves that identity [3] holds indeed.

********

There are several other proofs of Morley's theorem. One was given by
Professor John H. Conway of Princeton in the geometry-puzzles
newsgroup, hosted by the Math Forum. Note that he starts with a
triangle with angles 3A, 3B and 3C, so that A, B and C are the
trisected angles.

The message can be found at:

http://mathforum.org/kb/message.jspa?messageID=1084185

(Quoting)

This is "Morley's Trisector Theorem", which has long had a justly
deserved reputation as being difficult to prove. So I was very
pleased when I found a very simple proof some years ago. Here it
is: Let the angles be  3A, 3B, 3C, and let  "X+" mean  "X + 60
degrees". Then there certainly exist 7 abstract triangles having
the angles:

A++,B,C; A,B++,C; A,B,C++; A,B+,C+; A+,B,C+; A+,B+,C; 0+,0+,0+
1       2       3         4        5        6        7

since in every case the triple of angles adds to 180 degrees. Now
these triangles are only determined up to scale, and I determine
the scale by saying that certain lines are all to have the same
length.

I draw the situation below, but unfortunately can't draw all the
lines and can't put the vertices exactly where I want them:

B    B    B

5

3
C++   A+_____C+   A++ 1

0+_______0+
\     /
B+    \ 7 /    B+
A           \     \ /    /            C
\     0+   /
4   \        /   6
C+     A+

A             B++              C
/ \
/ 2 \
/     \
A--------X-------Y----------C

Triangle number 7, with angles 0+,0+,0+, is clearly equilateral,
and can take all its edges to have some fixed length L. Then I
arrange that the edges

B+ C+     C+ A+     A+ B+

of triangles 4, 5, 6 should also have length L. The way we choose
the scale of the other three triangles I illustrate only with
triangle number 2. Namely, I draw the isosceles triangle B++ X Y
whose base angles have the value B+, and make the two slanting
sides of this have length L.

Then it's easy to see that these all fit together to make up a
triangle whose angles are 3A, 3B, 3C, and which is therefore
similar to the original one, so proving Morley's theorem. To see
this, you just have to check that any two sides that come together
have the same length, and that the angles around any internal
vertex add to 360 degrees. The latter is easy, and the former is
proved using congruences such as that that takes the vertices A,
C+, B+  of triangle number 4 to the points A, B++, Y  of triangle
number 2.

I know of no other proof that's less than twice as long!

John Conway

********

It can be shown that by a good choice of external angle trisectors
instead of the internal ones we used above, there can be found 17 more
equilateral trisector triangles.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 08/14/2000 at 05:27:00

Hi, Dr. Floor,

solved without reference to trigonometry? If so, I would appreciated

Thanks a lot.
```

```
Date: 08/15/2000 at 06:46:39
From: Doctor Floor

Thanks for your reaction. I thought that perhaps the "puzzle proof" by
John Conway was such a proof without trigonometry.

Here is another proof that, not unlike Conway's proof, works somewhat
the other way around:

Let DEF be an equilateral triangle. Let P be a point on the extended
median from D (the line connecting D and the midpoint of EF), beyond
the midpoint of EF. Points Q and R are chosen likewise on the medians
from E and F respectively. Let A be the point of intersection of RE
and QF, B of PF and RD and C of PE and QD.

Now let l be the reflection of the line BP through BR, and m the
reflection of CP through CQ; and let X be the intersection of l and m.
We can define the points Y and Z likewise.

Note that AR and AQ are the trisectors of <YAZ. We will name the
magnitude of <RAQ = a. Similarly <PBR = b and <QCP = c.

Let x be the base angle of (isosceles) triangle PEF, y of QFD and z of
RDE. Then

<BDC = < QDR = y + z + 60 deg

and in the same way

<CEA = x + z + 60 deg
<AFB = x + y + 60 deg

Now note that

<AED = <AER - <RED = 180 deg - z

and in the same way

<AFD = 180 deg - y

Since the angles of a quadrilateral add to 360 deg, we find in AFDE

a + 60 deg  + (180 deg - y) + (180 deg - z) = 360 deg
a + 60 deg = y + z
a = y + z - 60 deg    ....................................[1]

With this we can compute

<AZB = 360 deg - a - b - <AFB
= 360 deg - a - b - (x + y + 60 deg)
= 240 deg - (a + b + c)   ...........................[2]

From the symmetry of this result, we see that <BXC and <CYA must have
the same value.

Why all this reasoning?

Well, if we let a', b', c' be given so that a' + b' + c' = 60 deg, so
that a', b', and c' are the trisected angles of a given triangle. Then
we build the above construction with x = 60 deg - a', y = 60 deg - b'
and z = 60 deg - c'. The result we get is that:

* From [1] we see

a = y + z - 60 deg = 60 deg - b' - c' = a'

and in the same way

b = b'
c = c'

* From [2] we see that

<AZB = 240 deg - 60 deg = 180 deg

so Z is a straight angle (and so are angles CYA and BXC).

And we conclude that ABC is a triangle with the same angles as the
given triangle, and that the angle trisectors intercept an equilateral
triangle. As desired, to prove Morley's theorem.

I learned this proof from a booklet by O. Bottema ("Hoofdstukken uit
de Elementaire Meetkunde").

I hope this is a proof you like even better.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 08/22/2000 at 05:54:57

Hi again, Dr. Floor

reference to trigonometry.

x = 60 deg - a'
y = 60 deg - b'
z = 60 deg - c'

and identify a', b', and c' in the figure.

Best regards,
```

```
Date: 08/22/2000 at 09:35:01
From: Doctor Floor

In my message, until the line "Why all this reasoning?" I did
preliminary work. The figure is just for this first part. After this
part a new episode begins, in which I start of course with a triangle.

That triangle has angles, say, A', B', and C' (I already used A, B,
and C). After that I take a' = A/3, b' = B/3, c' = C/3, and since
A+B+C = 180 deg, we have a'+b'+c'= 60 deg.

I then TAKE x = 60 deg - a', y = 60 deg - b' and z = 60 deg - c' and
use these specific x, y, and z in the more general case of the first
part. You should think that all the work of the first part is redone.
So x, y, and z are not calculated, they are taken such that we get the
results we want (that is, these x, y, and z are chosen such that the
A, B, and C from the first part form a triangle similar to A'B'C', and
thus proving Morley's theorem). And indeed that works out great.

I hope this clears it up.

Kind regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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