Question Number 216033 by efronzo1 last updated on 26/Jan/25
Answered by mr W last updated on 26/Jan/25
Commented by mr W last updated on 26/Jan/25
$${AE}={BE}=\frac{{BC}}{\mathrm{2}} \\ $$$${AD}×{BE}^{\mathrm{2}} +{DB}×{AE}^{\mathrm{2}} ={AB}×\left({DE}^{\mathrm{2}} +{AD}×{BD}\right) \\ $$$$\left({AD}+{DB}\right)×\frac{{BC}^{\mathrm{2}} }{\mathrm{4}}={AB}×\left({DE}^{\mathrm{2}} +{AD}×{BD}\right) \\ $$$$\frac{{BC}^{\mathrm{2}} }{\mathrm{4}}={DE}^{\mathrm{2}} +{AD}×{BD} \\ $$$$\Rightarrow\frac{{BC}^{\mathrm{2}} }{{AD}×{BD}+{DE}^{\mathrm{2}} }=\mathrm{4}\:\checkmark \\ $$
Answered by som(math1967) last updated on 26/Jan/25
Commented by som(math1967) last updated on 26/Jan/25
$${EL}\bot{AB} \\ $$$$\therefore{EL}=\frac{\mathrm{1}}{\mathrm{2}}{AC}\:,{BL}={AL}=\frac{\mathrm{1}}{\mathrm{2}}{AB} \\ $$$$\:{DE}^{\mathrm{2}} ={DL}^{\mathrm{2}} +{EL}^{\mathrm{2}} \\ $$$$\Rightarrow{DE}^{\mathrm{2}} =\left({BD}−{BL}\right)^{\mathrm{2}} +\frac{{AC}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow{DE}^{\mathrm{2}} ={BD}^{\mathrm{2}} +{BL}^{\mathrm{2}} −\mathrm{2}{BD}.{BL}+\frac{{AC}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow{DE}^{\mathrm{2}} ={BD}^{\mathrm{2}} −{BD}.{AB}+\frac{{AC}^{\mathrm{2}} +{AB}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow{DE}^{\mathrm{2}} +{BD}\left({AB}−{BD}\right)=\frac{{BC}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow{DE}^{\mathrm{2}} +{BD}.{AD}=\frac{{BC}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\therefore\:\frac{{BC}^{\mathrm{2}} }{{DE}^{\mathrm{2}} +{BDAD}}=\mathrm{4} \\ $$