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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-1-4-a-2-b-2-c-2-e-2-f-2-




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...  (1/4)(a^2 +b^2 +c^2 +e^2 +f^2 )+4PE^2 =x^2 +y^2 +z^2 +t^2
$${P}-{intersection}\:{point}\:{of}\:{bimedians}\:{in} \\ $$$${ABCD}-{convexe}\:{quadrilateral}\:{with} \\ $$$${a};{b};{c};{d}-{sides},\:{e};{f}-{diagonals}, \\ $$$${E}-{point}\:{in}\:{plane},\:{x};{y};{z};{t}-{distances} \\ $$$${from},\:{E}-{to}\:{A};{B};{C};{D}.\:{Prove}\:{that}… \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \right)+\mathrm{4}{PE}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{t}^{\mathrm{2}} \\ $$
Commented by mathdanisur last updated on 12/May/21
please mr W Sir...
$${please}\:\boldsymbol{{mr}}\:\boldsymbol{{W}}\:{Sir}… \\ $$
Commented by mr W last updated on 13/May/21
easy to prove with vector method.  see Q140838
$${easy}\:{to}\:{prove}\:{with}\:{vector}\:{method}. \\ $$$${see}\:{Q}\mathrm{140838} \\ $$

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