Sine Law
a/sin(A) = b/sin(B) = c/sin(C) = 2r, if r=0.5, c=
sin C; b=sin B; a=sinA.
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PROOF Bankoff starts
out with a large triangle ABC, with angle trisectors drawn at points
A, B and C. From the above figure we know that A+B+C = 180^{o}.
Note that ADC
= 180 A/3  C/3
= 180  (A+C)/3
= 180  (180B)/3
= 120+ B/3
= (360 +B)/3
sin ADC
=sin((360+B)/3) =sin((180B)/3) (i)
For simplification, let A = 3,
B= 3, C= 3.
this implies ++
=60^{o}. if we assume that the radius of the circle circumscribed
around ABC
= 1 (i.e. r=1).
using sine law we get
AB = 2sinC, AB= 2sin(3);
BC= 2sinA, BC= 2sin(3);
AC= 2sinB, AC= 2sin(3).
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If we apply sine law to
BFC, we
have
BF/ sin = BC
/ sin (180).
by substitution:
BC/ sin (180)
= 2sin(3)/sin(+)
= 2sin(3)
/ sin(60).
Therefore BF/ sin
= 2sin(3)/
sin(60),
i.e. BF = 2sin(3)
sin c / sin (60).
Using the identity sin(3x) = 3sin(x)  4sin^{3}(x),
we can simplify the identity.
sin(3x)
= 4sin(x) [ (¡Ô3/2)^{2}sin^{2}(x)]
= 4sin(x) [ sin^{2}( 60)  sin^{2}(x)]
= 4sin(x) (sin 60 + sin(x)) (sin 60  sin (x))
= 4sin(x) 2 sin[(60+x)/2] cos [(60x)/2] 2sin[(60x)/2] cos [(60+x)/2]
= 4sin(x) sin (60+x)sin(60x).
therefore
BF= 8 sin()
sin (60+ )sin(60)sin
()/ sin(60)
BF= 8 sin()
sin (60+) sin
() (ii)
Applying the Sine Law,
AD/ sin()= AC/
sin((180B)/3) = 2r.
Recall from above that with the assumption that the circumradius r
= 0.5, we have AC= b= sin(B). Also =C/3.
Therefore, we have
AD*sin((180B)/3 = 2r sin(B) sin(C/3)
If we do similar work as in (ii), using B=3
and C=3, we
will get
AD = 8r sin()
sin() sin(60+
), and
AE = 8r sin()
sin() sin((60+).
Note that the ratio
AE/AD = sin(60+)/sin(60+).
But ADE
+ AED =
180  A/3
= (540A)/3
= (540(180BC))/3
= (360+B+C)/3
= (180+B)/3 + (180+C)/3. From here,
ADE = (180+B)/3
and AED
= (180 + C)/3,
and similarly for triangles BFE and DFC.
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It thus follows that
the sum of angles around F, excluding DFE is 300^{o}, or DFE
= 60^{o}. The other two angles are similarly shown to be 60^{o}.
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