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 Equality in pentagon (Posted on 2015-05-03)
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
.

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 Solution Comment 1 of 1

`Let f and s denote the lengths of the diagonalsand 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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