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May 21st, 2018, 10:31 PM   #1
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Prove that points are coplanar

Let $\displaystyle ABCDA_{1}B_{1}C_{1}D_{1}$ be the cube with volume of $\displaystyle 1000cm^{3}$. Let $\displaystyle E$, $\displaystyle F$, $\displaystyle G$, $\displaystyle H$, $\displaystyle I$ and $\displaystyle J$ be the middle points of edges $\displaystyle BC$, $\displaystyle CD$, $\displaystyle DD_{1}$, $\displaystyle D_{1}A_{1}$, $\displaystyle A_{1}B_{1}$ and $\displaystyle B_{1}B$, respectively. Prove that points $\displaystyle E$, $\displaystyle F$, $\displaystyle G$, $\displaystyle H$, $\displaystyle I$ and $\displaystyle J$ belong to the same plane (are coplanar), and calculate surface area and volume of the pyramid $\displaystyle AEFGHIJ$.
I managed to calculate surface area and volume, but my question is about proof of coplanarity. How to prove that (without using vectors)? What idea is valid (in which direction to go):
- that angle between two planes is constant;
- that diagonals of the obtained regular hexagon are of the same length as are in 2D;
- or something else?
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May 22nd, 2018, 08:46 AM   #2
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If you establish a coordinate system with a corner at $\displaystyle (0,0,0)$ then you can get coordinates $\displaystyle E(10,5,0)$, $\displaystyle F(5,10,0)$, $\displaystyle G(0,10,5)$, $\displaystyle H(0,5,10)$, $\displaystyle I(5,0,10)$, $\displaystyle J(10,0,5)$. All of which satisfy the plane $\displaystyle x+y+z=15$.
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