等同于證AC*AD=AB*(AC+AD)

設(shè)圓的半徑為1

所以

AB=2sin(π/7)

AC=2sin(2π/7)

AD=2sin(3π/7)

AC*AD

=4sin(2π/7)sin(3π/7)

=(-2)(cos(5π/7)-cos(π/7))

AB*(AC+AD)

=4sin(π/7)*(sin(2π/7)+sin(3π/7))

=(-2)(cos(3π/7)-cos(π/7)+cos(4π/7)-cos(2π/7))

=(-2)(cos(π-(4π/7))-cos(π/7)+cos(4π/7)-cos(π-(5π/7)))

=(-2)(cos(5π/7)-cos(π/7))

=AC*AD

故命題得證