Question Number 140785 by mathdanisur last updated on 12/May/21
$${P}-{intersection}\:{point}\:{of}\:{bimedians}\:{in} \\ $$$${ABCD}-{convexe}\:{quadrilateral}\:{with} \\ $$$${a};{b};{c};{d}-{sides},\:{e};{f}-{diagonals}, \\ $$$${E}-{point}\:{in}\:{plane},\:{x};{y};{z};{t}-{distances} \\ $$$${from},\:{E}-{to}\:{A};{B};{C};{D}.\:{Prove}\:{that}... \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \right)+\mathrm{4}{PE}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{t}^{\mathrm{2}} \\ $$
Commented by mathdanisur last updated on 12/May/21
$${please}\:\boldsymbol{{mr}}\:\boldsymbol{{W}}\:{Sir}... \\ $$
Commented by mr W last updated on 13/May/21
$${easy}\:{to}\:{prove}\:{with}\:{vector}\:{method}. \\ $$$${see}\:{Q}\mathrm{140838} \\ $$