Let each side of the regular n-sided polygon be #a# and the radius of the circumcircle of the polygon be R. Each side of the polygon will subtend angle #theta=(2pi)/n# at the center #O# of the circumcircle.
The perpendicular dropped from #O# to #A_1A_2# will bisect the #angleA_1OA_2=theta# as well as side #A_1A_2#
So #(A_1A_2)/2=Rsin(theta/2)#
Similarly the perpendicular dropped from #O# to #A_1A_3# will bisect the #angleA_1OA_3=2theta# as well as side #A_1A_3#
So #(A_1A_3)/2=Rsin((2theta)/2)#
And also the perpendicular dropped from #O# to #A_1A_4# will bisect the #angleA_1OA_4=3theta# as well as side #A_1A_4#
So #(A_1A_4)/2=Rsin((3theta)/2)#
Now given condition is
#1/(A_1A_2)=1/(A_1A_3)+1/(A_1A_4)#
Substituting the values of
#A_1A_2,A_1A_3andA_1A_4# we have
#1/(2Rsin(theta/2))=1/(2Rsin((2theta)/2))+1/(2Rsin((3theta)/2))#
#>1/(sin((2theta)/2))=1/(sin(theta/2))-1/(sin((3theta)/2))#
#=>1/(sin((2theta)/2))=(sin((3theta)/2)-sin(theta/2))/(sin((3theta)/2)sin(theta/2))#
#=>1/(sin((2theta)/2))=(2cos((2theta)/2)sin(theta/2))/(sin((3theta)/2)sin(theta/2))#
#=>sin((3theta)/2)=2cos((2theta)/2)sin((2theta)/2)#
#=>sin((3theta)/2)=sin((4theta)/2)#
#=>sin(pi-(3theta)/2)=sin((4theta)/2)#
#=>pi-(3theta)/2=(4theta)/2#
#=>(7theta)/2=pi#
#=>7/2*(2pi)/n=pi#
#=>n=7#
