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


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Question Number 140785 by mathdanisur last updated on 12/May/21

$P - i n t e r s e c t i o n p o i n t o f b i m e d i a n s i n$ $A B C D - c o n v e x e q u a d r i l a t e r a l w i t h$ $a; b; c; d - s i d e s, e; f - d i a g o n a l s,$ $E - p o i n t i n p l a n e, x; y; z; t - d i s t a n c e s$ $f r o m, E - t o A; B; C; D . P r o v e t h a t . . .$ $\frac{1}{4} (a^{2} + b^{2} + c^{2} + e^{2} + f^{2}) + 4 {P E}^{2} = x^{2} + y^{2} + z^{2} + t^{2}$
Commented by mathdanisur last updated on 12/May/21

$p l e a s e m r W S i r . . .$
Commented by mr W last updated on 13/May/21

$e a s y t o p r o v e w i t h v e c t o r m e t h o d .$ $s e e Q 140838$
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