Question

Se considera paralelipipedul dreptunghic ABCDA'B'C'D' si {O}=AB'∩A'B. Pe muchia [BC] se considera un punct N astfel incat A'C||(B'AN). Stiind ca D'O⊥(B'AN) demonstrati ca ABCDA'B'C'D'
este cub.

Answer 1

$(B'AN) \bigcap (A'BC)=ON ~si~ A'C \subset (A'BC),~A'C~|| ~(B'AN) \Rightarrow \\ A'C ~||~ON,~dar~O-mijlocul~[A'B]~\Rightarrow~M-mijlocul~lui~[BC]. \\ \\ D'O~ \bot~(B'AN) \Rightarrow D'O ~\bot~A'B',~O-mijlocul~lui~[A'B'] \Rightarrow \\ \Delta~D'AB'-isoscel~\Rightarrow~B'D'=AD'. \\ \\ Notez~AB=a,~BC=b~si~AA'=c. \\ \\ B'D'^2=AD'^2 \Leftrightarrow b^2+c^2=a^2+c^2 \Rightarrow b^2=a^2 \Rightarrow a=b. \\ \\ D'O ~\bot~ON \Rightarrow~D'O^2+ON^2=D'N^2.~(*)$

$Fie~M \in A'B'~si~P \in AB ~a.i~MP ~||~AA'~si~M,~O,~P-coliniare. \\ \\ D'O^2=D'M^2+MO^2=A'D'^2+A'M^2+MO^2= a^2+\frac{a^2}{4}+ \frac{c^2}{4} = \\ = \frac{ 5a^{2}+c^ 2}{4} . \\ \\ ON^2=OP^2+PN^2= \frac{c^2}{4}+ \frac{a^2}{2}= \frac{2a^2+c^2}{4} . \\ \\ D'N^2=DD'^2+DN^2= c^{2}+ \frac{5a^2}{4}= \frac{5a^2+c^2}{4}. \\ \\ (*) \Leftrightarrow \frac{5a^2+c^2}{4}+ \frac{2a^2+c^2}{4}= \frac{4c^2+5a^2}{4} \Leftrightarrow 5a^2+c^2+2a^2+c^2=4c^2+5a^2 \Leftrightarrow \\ \Leftrightarrow 7a^2+2c^2=4c^2+5a^2$
$\Leftrightarrow 2a^2=2c^2 \Rightarrow a=c. \\ \\ a=b=c \Rightarrow cub.$