From the above figure
we know that 3a+3b+3c = 180^{o}, which means that a+b+c= 60^{o}.
Let x^{+} = x+60^{o} for any angle
x. since a+b+c = 60^{o}. Let 0^{+}=a+b+c=60^{o}.,
This shows that it is possible to construct an triangle with 3 different
types of angle combinations:
Type 1: 
0^{+},0^{+},0^{+}; 
Type 2: 
a,b^{+},c^{+}; 
a^{+},b,c^{+}; 
a^{+},b^{+},c 
Type 3: 
a^{++},b,c; 
a,b^{++},c; 
a,b,c^{++} 
since these seven combinations of angles all have
a sum of 180^{o}.
Instead of working forward, Conway worked backwards.
He showed that from an equilateral triangle one can construct a
triangle with any angles, i.e. with arbitrary a,b and c (that sums
to 60^{o}). According to Conway, we can make the following
constructions:
Type 1. Construct an equilateral triangle with length
1.
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ps file
Type 2. Construct a triangle with the side joining
larger angles ( e.g. a^{+} and b^{+}) to have length
1. for example:
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Type 3. Construct two lines that intersects the
side opposite to b^{++} at angle b^{+}, thus forming
an isosceles triangle with base angle =b^{+}.
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ps file
The significance of constructing the isosceles triangle
is to prove JHI
and DFC
are congruent.
1.JHI
= DFC
= b^{+}
2. JIH
= DCF
= c
3. By construction, we get JH = DF=1
The above proves that JHI
andDFC
are congruent. This result is important because it shows that DC
= JI are equal, when the two triangles are matched together, they
become the common edge, then point J = point D and point I = point
C.
Other triangles are constructed in similar way,
and we can get the following:
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Note that the angles at the vertex of the equilateral
triangle (e.g. D) sums up to 360^{o}.
Check: (a^{+})+(c^{+})+(b^{++})+(0^{+})
= a+60^{o}+c+60^{o}+b+120^{o}+60^{o}
=(a+b+c)+300^{o}=360^{o}.
Since the angles at D sums up to 360^{o},
and since the sides match due to congruency, we see that the
triangles can be assembled perfectly together to form a larger triangle
with angles 3a, 3b and 3c, and hence the conclusion.
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to enlarge
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