Download ps file

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.

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).

Download ps file

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.

sin(3x)
= 4sin(x) [ (¡Ô3/2)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).

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((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.

Click to enlarge Download ps file

Click to enlarge Download ps file