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Heptatriangle (Posted on 2014-03-01) Difficulty: 3 of 5

  
Let the interior angles of ΔABC satisfy the following:

   m(∠A) = 2*m(∠B) = 4*m(∠C).

If the bisectors of interior angles A, B, and C intersect
the opposite sides in points D, E, and F respectively,
prove that ΔDEF is isosceles.
  

  Submitted by Bractals    
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Solution: (Hide)

  
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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