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The proof below is a direct proof based on
the work of Bankoff. This proof uses less
geometry than the previous two. However it
requires some knowledge in trigonometry.

According to Alexander Bogomolny, this proof is first published
in Mathematics Magazine, 35 (1962) 223-224.<ref 2>

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 = 180o.

Note that ADC
= 180- A/3 - C/3
= 180 - (A+C)/3
= 180 - (180-B)/3
= 120+ B/3
= (360 +B)/3

sin ADC =sin((360+B)/3) =sin((180-B)/3)--- (i)

For simplification, let A = 3, B= 3, C= 3. this implies ++ =60o. 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) - 4sin3(x), we can simplify the identity.

= 4sin(x) [ (3/2)2-sin2(x)]
= 4sin(x) [ sin2( 60) - sin2(x)]
= 4sin(x) (sin 60 + sin(x)) (sin 60 - sin (x))
= 4sin(x) 2 sin[(60+x)/2] cos [(60-x)/2] 2sin[(60-x)/2] cos [(60+x)/2]
= 4sin(x) sin (60+x)sin(60-x).

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((180-B)/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((180-B)/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
= (540-A)/3
= (540-(180-B-C))/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 300o, or DFE = 60o. The other two angles are similarly shown to be 60o.

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