Geometry – Solution Verification for Proof of Cube and Prism

Question

question

The prism $ABCDA'B'C'D'$ is straight. We denote the means of the edges $(AB)$, $(BC)$ respectively $(BB')$ by $M$, $N$, $P $and the centers of the faces $ADD'A'$ , $DCC'D'$ respectively $A'B'C'D'$ with $M'$, $N'$ respectively $P'$. Prove that the triangular prism $MNPM'N'P'$ is regular if and only if the parallelepiped $ABCDA'B'C'D'$ is a cube.

Drawing

my idea

I posted this because I want to ask you if can you please tell me if my idea is right or wrong. Thank you so much!

1. We know that $ABCDA'B'C'D'$ is a cube

We show that $MNPM'N'P'$ is a regular prism

$M'N', P'N'$ and $M'P'$ are midlines in triangles $D'AC,C'A'D$ and $D'B'A$, which means that

$M'N'=\frac{AC}{2}$

$P'N'=\frac{A'D}{2}$

$M'P'=\frac{AB'}{2}$

also because $ABCDA'B'C'D'$ is a cube by Pythagoras we can say that $AB'=AC=A'D$

which means that $M'N'=P'N'=M'P'=> M'N'P'$ is equilateral

$MN, PM$ and $PN$ are midlines in triangles $ABC,ABB'$ and $BB'C$, which means that

$MN=\frac{AC}{2}$

$PN=\frac{B'C}{2}$

$MP=\frac{AB'}{2}$

also because $ABCDA'B'C'D'$ is a cube by Pythagoras we can say that $AB'=AC=B'C$

which means that $MN=PM=NP=>MNP$ is equilateral

by the 2 equilateral triangles, we can say that $MNPM'N'P'$ is a regular prism

1. We showthat $ABCDA'B'C'D'$ is a cube

We know that $MNPM'N'P'$ is a regular prism

Because $MNPM'N'P'$ is a regular prism, $M'N'=P'N'=M'P'$

$M'N', P'N'$ and $M'P'$ are midlines in triangles $D'AC,C'A'D$ and $D'B'A$, which means that

$M'N'=\frac{AC}{2}$

$P'N'=\frac{A'D}{2}$

$M'P'=\frac{AB'}{2}$

and because they are congruent we can show that $AC=AB'=A'D$

which we can write as $AC^2=AB'^2=A'D^2$

$h^2+l^2=h^2+L^2=L^2+l^2$ where h,l,L= are the dimensions of the parallelepiped

$h=l=L$ which means $ABCDA'B'C'D'$ is a cube

Answer 1

I think your proof is correct.

Your proof can be written in the following way. The following way might be simpler.

Considering $\triangle{BAC}$ and $\triangle{D'AC}$, we have $$MN=\frac 12AC=M'N'\tag1$$ Considering $\triangle{BB'C}$ and $\triangle{C'A'D}$, we have $$PN=\frac 12B'C=\frac 12A'D=P'N'\tag2$$ Considering $\triangle{BAB'}$ and $\triangle{D'B'A}$, we have $$PM=\frac 12AB'=P'M'\tag3$$

• If $ABCDA'B'C'D'$ is a cube, then we have $AC=B'C=AB'$. It follows from $(1)(2)(3)$ that $MN=M'N'=PN=P'N'=PM=P'M'$. So, the triangular prism $MNPM'N'P'$ is regular.

• If the triangular prism $MNPM'N'P'$ is regular, then $MN=PN=PM$. It follows from $(1)(2)(3)$ that $AC=B'C=AB'$. Since $AC^2=B'C^2=AB'^2$, we have $AB^2+BC^2=BC^2+BB'^2=AB^2+BB'^2$. Since $AB=BB'=BC$, we can say that $ABCDA'B'C'D'$ is a cube.