Let a, b, and c be the lengths of the sides opposite the
vertices A, B, and C respectively. Let x = m(∠C).
Apply the Law of Sines to ΔABC:
a b c
--------- = --------- = --------
sin(4x) sin(2x) sin(x)
b*sin(x) = c*sin(2x) = 2*c*sin(x)cos(x) ⇒
cos(∠C) = cos(x) = b/(2c) (1)
a*sin(2x) = b*sin(4x) = 2*b*sin(2x)cos(2x) ⇒
cos(∠B) = cos(2x) = a/(2b) (2)
a/(2b) = cos(2x) = 2*cos(x)2 - 1 ⇒
b3 = c2*(a + 2b) (3)
The Angle Bisector Theorem gives
|AE| = bc/(a + c) (4)
|AF| = bc/(a + b) (5)
ΔABC ~ ΔAEB ⇒
|AB|2 = |AC||AE|
c2 = b[bc/(a + c)]
b2 = c(a + c) (6)
Combining (3)&(6) gives
ab = c(a + b) (7)
ΔABC ~ ΔDAC ⇒
|AD| = |AB||AC|/|BC| = bc/a (8)
Apply the Law of Cosines to ΔDAE and ΔDAF:
|DE| = |DF|
⇔ |DE|2 = |DF|2
⇔ |AD|2 + |AE|2 - 2|AD||AE|cos(∠DAE)
= |AD|2 + |AF|2 - 2|AD||AF|cos(∠DAF)
⇔ |AE|2 - |AF|2 = 2|AD|(|AE| - |AF|)cos(2x)
⇔ |AE| + |AF| = 2|AD|cos(2x)
⇔ [bc/(a + c)] + [bc/(a + b)]
= 2*[bc/a]*[a/(2b)] = c [See 2,4,5,8]
⇔ b(a + b) = a(a + c)
⇔ b(a + b) = a[b2/c] [See 6]
⇔ ab = c(a + b) [See 7]
QED
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