The proof
uses the following lemma:
LEMMA Let BAC
= a, point D is the incentre of ABC
if and only if (a) the line DA bisects a and (b) CDB
= 90 + a/2.
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PROOF Suppose that angles ,
and
are given with ++
= 180. Dergiades' approach is to construct a triangle with these
angles for which Morley's Theorem is true.
According to Dergiades, suppose we let
x=60-(/3),
y=60-(/3),
z=60-(/3).
Then 0<x,y,z<60 and x+y+z=120.
Consider an equilateral triangle. Construct 3 isoceles
triangles with base angles x, y, z respectively ( with the constraints
x+y+z =120 and 0<x,y,z<60) on the sides on the equilateral
triangle.
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Extend the sides of the isosceles
triangles so that they intersect as shown:
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Since D is the vertex
of the equilateral triangle, and I is the vertex of the isosceles
triangle with base angle x, it follows that the line ID is the angle
bisector.
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Join B and C, note that BDC
= 60+z+y (vertically opposite angles)
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BDC
= 60+z+y = 60+x+y+z-x=180-x. Since BIC
= 180-2x. and D is bisects BIC,
the Lemma shows that D is the incentre of CIB.
It follows that the lines drawn at points A B and
C are angle trisectors.
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The last step is to prove
ABC has
angles ,
and . In the
following figure:
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we can see that
a+s+t+x+x= 180
s+x+z+60=180, i.e. s+x+z=120
t+x+y+60=180, i.e. t+x+y=120
x+y+z=120(given)
by substitution, we have
a+(s+x)+(t+x)=180
a+(120-z)+(120-y)=180
a+120+(120-z-y)=180
a+120+x=180
It follows that:
a=60-x= /3.
Hence ABC
has angles ,
and . |