Question Number 192288 by universe last updated on 14/May/23
Answered by mahdipoor last updated on 14/May/23
$${get}\:\angle{BCG}={a}\:\:\:\:\:,\:{BC}={b} \\ $$$${GD}={GB}+{BD}={BC}.{tana}+{BC}.{tan}\left(\mathrm{90}−{a}\right) \\ $$$$={b}\left({tana}+{cota}\right) \\ $$$${FD}={GD}.{tana}={b}\left({tan}^{\mathrm{2}} {a}+\mathrm{1}\right)={b}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {a}}\right) \\ $$$${EC}={CD}+{DE}=\frac{{BC}}{{sina}}+{FD}.{sina}={b}\left(\frac{\mathrm{1}}{{sina}}+\frac{{sina}}{{cos}^{\mathrm{2}} {a}}\right) \\ $$$${AB}={AC}−{BC}=\frac{{EC}}{{sina}}−{b}={b}\left(\frac{\mathrm{1}}{{sin}^{\mathrm{2}} {a}}+\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {a}}−\mathrm{1}\right) \\ $$$$\zeta=\frac{{AB}}{{BC}}=\frac{{cos}^{\mathrm{2}} {a}+{sin}^{\mathrm{2}} {a}−{sin}^{\mathrm{2}} {a}.{cos}^{\mathrm{2}} {a}}{{sin}^{\mathrm{2}} {a}.{cos}^{\mathrm{2}} {a}}= \\ $$$$\frac{\mathrm{1}−\frac{{sin}^{\mathrm{2}} \mathrm{2}{a}}{\mathrm{4}}}{\frac{{sin}^{\mathrm{2}} \mathrm{2}{a}}{\mathrm{4}}}=\frac{\mathrm{4}−{sin}^{\mathrm{2}} \mathrm{2}{a}}{{sin}^{\mathrm{2}} \mathrm{2}{a}} \\ $$$${min}\:\zeta=\zeta\left({sin}^{\mathrm{2}} \mathrm{2}{a}=\mathrm{1}\right)=\mathrm{3} \\ $$