Question Number 184511 by HeferH last updated on 07/Jan/23
$${In}\:{a}\:{regular}\:{heptagon}\:{ABCDEFG}\:: \\ $$$$\:\sqrt{\mathrm{2}\left({AC}\right)^{\mathrm{2}} −\left({AD}\right)^{\mathrm{2}} }\:−\:{AD}\:=\:\mathrm{2} \\ $$$$\:{find}\:{BC}.\: \\ $$