Answer
The line connecting the midpoints of two sides of a triangle is parallel to the third side.
Work Step by Step
Let $A$, $B$ and $C$ be the vertices of the triangle $\vartriangle ABC$. Suppose the components of the points are given by $A = \left( {{a_1},{a_2},{a_3}} \right)$, $B = \left( {{b_1},{b_2},{b_3}} \right)$ and $C = \left( {{c_1},{c_2},{c_3}} \right)$.
Let $P$ and $Q$ denote the midpoints of the line segments $\overline {AB} $ and $\overline {BC} $, respectively. So,
$P = \left( {\frac{{{a_1} + {b_1}}}{2},\frac{{{a_2} + {b_2}}}{2},\frac{{{a_3} + {b_3}}}{2}} \right)$
and
$Q = \left( {\frac{{{b_1} + {c_1}}}{2},\frac{{{b_2} + {c_2}}}{2},\frac{{{b_3} + {c_3}}}{2}} \right)$
Thus, the vector $\overrightarrow {PQ} $ is
$\overrightarrow {PQ} = Q - P$
$\overrightarrow {PQ} = \left( {\frac{{{b_1} + {c_1}}}{2},\frac{{{b_2} + {c_2}}}{2},\frac{{{b_3} + {c_3}}}{2}} \right) - \left( {\frac{{{a_1} + {b_1}}}{2},\frac{{{a_2} + {b_2}}}{2},\frac{{{a_3} + {b_3}}}{2}} \right)$
$\overrightarrow {PQ} = \left( {\frac{{{c_1} - {a_1}}}{2},\frac{{{c_2} - {a_2}}}{2},\frac{{{c_3} - {a_3}}}{2}} \right)$
$\overrightarrow {PQ} = \frac{1}{2}\left( {{c_1} - {a_1},{c_2} - {a_2},{c_3} - {a_3}} \right)$
Next, we find the vector $\overrightarrow {AC} $.
$\overrightarrow {AC} = C - A = \left( {{c_1},{c_2},{c_3}} \right) - \left( {{a_1},{a_2},{a_3}} \right)$
$\overrightarrow {AC} = \left( {{c_1} - {a_1},{c_2} - {a_2},{c_3} - {a_3}} \right)$
Comparing $\overrightarrow {PQ} $ and $\overrightarrow {AC} $, we see that $\overrightarrow {PQ} = \frac{1}{2}\overrightarrow {AC} $. Because $\overrightarrow {PQ} $ is a constant multiple of $\overrightarrow {AC} $, $\overrightarrow {PQ} $ is parallel to $\overrightarrow {AC} $. Since $A$, $B$ and $C$ are arbitrary vertices of any triangle $\vartriangle ABC$, we conclude that the line connecting the midpoints of two sides of a triangle is parallel to the third side.
