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Equality in pentagon (Posted on 2015-05-03) Difficulty: 3 of 5
Imagine regular pentagon ABCDE inscribed in a circle.
If point F is on arc BC, denote:
FA=a; FB=b; FC=c; FD=d; FE=e

Prove that a + d = b + c + e
.

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.0000 (1 votes)

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Solution Solution Comment 1 of 1

Let f and s denote the lengths of the diagonals
and sides of the pentagon respectively.
The problem is solved using four instances of 
Ptolemy's theorem.

Cyclic quadrilateral ABFE:
   |AF||BE| = |AB||EF| + |AE||BF|
     a * f  =   s * e  +   s * b
     a = s*(b + e)/f                               (1)

Cyclic quadrilateral CDEF:

   |CE||DF| = |CD||EF| + |CF||DE|
     f * d  =   s * e  +   c * s
     d = s*(c + e)/f                               (2)

Combining (1) & (2)

     a + d = s*(b + C + 2*e)/f               (3)

Cyclic quadrilateral BFCE:

   |CE||DF| = |CD||EF| + |CF||DE|
     s * e  =   f * c  +   b * f
     e = f*(b + c)/s                               (4)

Cyclic quadrilateral ABCD:

   |AC||BD| = |AB||CD| + |AD||BC|
     f * f  =   s * s  +   f * s
     s*f - s^2 = 2*s*f - f^2
     s*(f - s) = f*(2*s - f)

       f - s      f
     --------- = ---  and  (4)  ==>
      2*s - f     s

           f - s      
     e = --------- * (b + c)
          2*s - f  

     2*s*e - e*f = f*(b + c) - s*(b + c)
     s*(b + c + 2*e) = f*(b + c + e)
     s*(b + c + 2*e)/f = b + c + e   and  (3)  ==>

     a + d = b + c + e
Note: The endpoints B and C must be included
in the arc BC.

QED             
        

Edited on May 3, 2015, 12:59 pm

Edited on May 7, 2015, 12:38 am
  Posted by Bractals on 2015-05-03 10:49:10

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