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