<s>[INST] Height of a cube edge from the floor 
A cube $ABCD.EFGH$ has side length $2a$ cm. Point $A$ is lifted $a$ cm from the floor, point $C$ is still on the floor, point $B$ and point $D$ are on the same height from the floor. What is the height of point $E$ from the floor?

I'm sorry for my bad English, but I try to illustrate it as follows.

The left cube is the original one while the right side is the cube after we lifted the point $A$. In my mind, to find the height of point $E$ from the floor is to find the length of line $EN$. We can use Pythagorean theorem $EN^2=ME^2-MN^2$. But how to find the length of line $ME$ and $MN$?
 [/INST] You can also use similar triangles. Let $P$ be the point on $MN$ directly below $A$, and let $Q$ be the point of intersection of $EN$ and line $AC$. Since $\angle CAM$ is a right angle, we have that
$$\triangle CAP \sim \triangle AMP \sim \triangle EAQ$$
$CA \,$ is clearly $\, 2a \sqrt{2}\, $ so $\, PC = a\sqrt{7}\,.$ Then
$$\begin{align}
\frac{AC}{PC} &= \frac{EA}{EQ} \\
\\
EQ &= \frac{EA \cdot PC}{AC} \\
\\
&= \frac{(2a)(a\sqrt{7})}{2a\sqrt{2}} \\
\\
&= a \sqrt {\frac{7}{2}} \\
\\
&= \frac{a}{2} \sqrt{14}
\end{align}$$
and so the height of $E$ is
$$\begin{align}
EQ + QN &= \frac{a}{2} \sqrt{14} + a \\
\\
&= \boxed{ a \left( \frac{\sqrt{14}}{2} + 1 \right) }
\end{align}$$
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