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
.
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
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Posted by Bractals
on 2015-05-03 10:49:10 |
Solution