Shifting of Line Segment                       
$L_1$ and $L_2$ are skew lines in $3$-dimensional space. 
$A_1 C \mathbin{\bot} L_1,L_2$. If line segment $AB$ is shifted parallelly, such that $A$ moves along $L_1$ and reaches $A_1$ while $B$ reaches $B_1$, then prove that $B_1 C \mathbin\bot A_1 C$.
It seems really intuitive but I wasn't able to get it. Can anyone prove it using only simple geometry (lines, congruency or similarity) without vectors?

Here's a vector solution, but coordinate-free, and really simple  . . .

The advantages? It works in $n$ dimensions, for all $n \ge 2$, and requires minimal visualization.

From $A_1 C \mathbin{\bot} L_1,L_2$, we get
$$(\mathbf{A_1}-\mathbf{C})\cdot(\mathbf{A}-\mathbf{A_1})=0\tag{eq1}$$
$$(\mathbf{A_1}-\mathbf{C})\cdot(\mathbf{B}-\mathbf{C})=0\tag{eq2}$$
Since line segment $AB$ moves in  parallel, ending up as line segment $A_1B_1$, it follows that
$$\mathbf{B}-\mathbf{A}=\mathbf{B_1}-\mathbf{A_1}\tag{eq3}$$
The goal is to prove $B_1 C \mathbin\bot A_1 C$, or equivalently
$$(\mathbf{B_1}-\mathbf{C})\cdot(\mathbf{A_1}-\mathbf{C})=0\tag{g}$$
Now here's the intuition  . . .

The $3$ equations $(\text{eq}1),\,(\text{eq}2),\,(\text{eq}3)$, represent all the information in the problem, effectively replacing the diagram. Hence, either those $3$ equations identically imply equation $(\text{g})$, or else the implication doesn't hold, in which case, the claim of the problem is false.

With that intuition, elementary algebra (of vectors) should suffice . . .

Subtracting $(\text{eq}1)$ from $(\text{eq}2)$, we get
\begin{align*}
&
(\mathbf{A_1}-\mathbf{C})
\cdot
\bigl((\mathbf{B}-\mathbf{C})
-
(\mathbf{A}-\mathbf{A_1})
\bigr)
=0
\\[4pt]
\implies\;&
(\mathbf{A_1}-\mathbf{C})
\cdot
\bigl((\mathbf{B}-\mathbf{A})
-
(\mathbf{C}-\mathbf{A_1})
\bigr)
=0
\\[4pt]
\implies\;&
(\mathbf{A_1}-\mathbf{C})
\cdot
\bigl((\mathbf{B_1}-\mathbf{A_1})
-
(\mathbf{C}-\mathbf{A_1})
\bigr)
=0
&&\text{[by $(\text{eq}3)$]}
\\[4pt]
\implies\;&
(\mathbf{A_1}-\mathbf{C})
\cdot
(\mathbf{B_1}-\mathbf{C})
=0
\\[4pt]
\end{align*}
which proves equation $(\text{g})$, as required.
