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Question Number 196829 by cortano12 last updated on 01/Sep/23

Answered by universe last updated on 01/Sep/23

by apollonius theorem BC^2 + PC^(2 ) = 2(QC^2 +PQ^2 ) .....(1) AC^2 + QC^2 = 2(PC^2 + PQ^2 ) .....(2) (1)+(2) AC^2 +BC^2 +PC^2 +QC^2 = 2QC^2 +2PC^2 +4PQ^2 AB^2 = 3PQ^2 +PQ^2 + QC^2 +PC^2 AB^2 −3PQ^2 = PQ^2 +PC^2 + QC^2 AB = 3PQ AB^2 −((3AB^2 )/9) = PQ^2 + PC^2 + QC^2 (2/3)AB^2 = PQ^2 + QC^2 + PC^2

$${by}\:{apollonius}\:{theorem} \\ $$$${BC}^{\mathrm{2}} \:+\:{PC}^{\mathrm{2}\:} \:=\:\mathrm{2}\left({QC}^{\mathrm{2}} +{PQ}^{\mathrm{2}} \right)\:\:\:.....\left(\mathrm{1}\right) \\ $$$${AC}^{\mathrm{2}} \:+\:{QC}^{\mathrm{2}} \:\:=\:\:\mathrm{2}\left({PC}^{\mathrm{2}} \:+\:{PQ}^{\mathrm{2}} \right)\:.....\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right) \\ $$$${AC}^{\mathrm{2}} +{BC}^{\mathrm{2}} +{PC}^{\mathrm{2}} +{QC}^{\mathrm{2}} \:=\:\mathrm{2}{QC}^{\mathrm{2}} +\mathrm{2}{PC}^{\mathrm{2}} +\mathrm{4}{PQ}^{\mathrm{2}} \\ $$$${AB}^{\mathrm{2}} \:=\:\mathrm{3}{PQ}^{\mathrm{2}} +{PQ}^{\mathrm{2}} \:+\:{QC}^{\mathrm{2}} +{PC}^{\mathrm{2}} \\ $$$${AB}^{\mathrm{2}} −\mathrm{3}{PQ}^{\mathrm{2}} \:=\:\:{PQ}^{\mathrm{2}} +{PC}^{\mathrm{2}} +\:{QC}^{\mathrm{2}} \\ $$$$\:\:{AB}\:=\:\mathrm{3}{PQ} \\ $$$${AB}^{\mathrm{2}} −\frac{\mathrm{3}{AB}^{\mathrm{2}} }{\mathrm{9}}\:=\:{PQ}^{\mathrm{2}} \:+\:{PC}^{\mathrm{2}} \:+\:{QC}^{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}{AB}^{\mathrm{2}} \:=\:{PQ}^{\mathrm{2}} \:+\:{QC}^{\mathrm{2}} \:+\:{PC}^{\mathrm{2}} \\ $$

Commented by cortano12 last updated on 01/Sep/23

AB^2 −((3AB^2 )/9) = (2/3)AB^2

$$\mathrm{AB}^{\mathrm{2}} −\frac{\mathrm{3AB}^{\mathrm{2}} }{\mathrm{9}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{AB}^{\mathrm{2}} \\ $$

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