Question Number 196829 by cortano12 last updated on 01/Sep/23
Answered by universe last updated on 01/Sep/23
$${by}\:{apollonius}\:{theorem} \\ $$$${BC}^{\mathrm{2}} \:+\:{PC}^{\mathrm{2}\:} \:=\:\mathrm{2}\left({QC}^{\mathrm{2}} +{PQ}^{\mathrm{2}} \right)\:\:\:…..\left(\mathrm{1}\right) \\ $$$${AC}^{\mathrm{2}} \:+\:{QC}^{\mathrm{2}} \:\:=\:\:\mathrm{2}\left({PC}^{\mathrm{2}} \:+\:{PQ}^{\mathrm{2}} \right)\:…..\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right) \\ $$$${AC}^{\mathrm{2}} +{BC}^{\mathrm{2}} +{PC}^{\mathrm{2}} +{QC}^{\mathrm{2}} \:=\:\mathrm{2}{QC}^{\mathrm{2}} +\mathrm{2}{PC}^{\mathrm{2}} +\mathrm{4}{PQ}^{\mathrm{2}} \\ $$$${AB}^{\mathrm{2}} \:=\:\mathrm{3}{PQ}^{\mathrm{2}} +{PQ}^{\mathrm{2}} \:+\:{QC}^{\mathrm{2}} +{PC}^{\mathrm{2}} \\ $$$${AB}^{\mathrm{2}} −\mathrm{3}{PQ}^{\mathrm{2}} \:=\:\:{PQ}^{\mathrm{2}} +{PC}^{\mathrm{2}} +\:{QC}^{\mathrm{2}} \\ $$$$\:\:{AB}\:=\:\mathrm{3}{PQ} \\ $$$${AB}^{\mathrm{2}} −\frac{\mathrm{3}{AB}^{\mathrm{2}} }{\mathrm{9}}\:=\:{PQ}^{\mathrm{2}} \:+\:{PC}^{\mathrm{2}} \:+\:{QC}^{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}{AB}^{\mathrm{2}} \:=\:{PQ}^{\mathrm{2}} \:+\:{QC}^{\mathrm{2}} \:+\:{PC}^{\mathrm{2}} \\ $$
Commented by cortano12 last updated on 01/Sep/23
$$\mathrm{AB}^{\mathrm{2}} −\frac{\mathrm{3AB}^{\mathrm{2}} }{\mathrm{9}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{AB}^{\mathrm{2}} \\ $$