Problem on geometric proofs with vectors
$\newcommand{\bfA}{\mathbf{A}}$ $\newcommand{\bfB}{\mathbf{B}}$ $\newcommand{\bfC}{\mathbf{C}}$ $\newcommand{\bfF}{\mathbf{F}}$ $\newcommand{\bfI}{\mathbf{I}}$ $\newcommand{\bfa}{\mathbf{a}}$ $\newcommand{\bfb}{\mathbf{b}}$ $\newcommand{\bfc}{\mathbf{c}}$ $\newcommand{\bfd}{\mathbf{d}}$ $\newcommand{\bfe}{\mathbf{e}}$ $\newcommand{\bfi}{\mathbf{i}}$ $\newcommand{\bfj}{\mathbf{j}}$ $\newcommand{\bfk}{\mathbf{k}}$ $\newcommand{\bfn}{\mathbf{n}}$ $\newcommand{\bfr}{\mathbf{r}}$ $\newcommand{\bfu}{\mathbf{u}}$ $\newcommand{\bfv}{\mathbf{v}}$ $\newcommand{\bfw}{\mathbf{w}}$ $\newcommand{\bfx}{\mathbf{x}}$ $\newcommand{\bfy}{\mathbf{y}}$ $\newcommand{\bfz}{\mathbf{z}}$
Show that the line connecting the midpoints of two sides of a triangle is parallel to and half the length of the third side.
Solution
Begin a geometric proof by labeling important points
We begin by labeling the triangle's vertices as $\bfa, \bfb,\bfc$.
Next, we express the two midpoints in terms of $\bfa, \bfb,\bfc$.
Recall that
Midpoint between two vectors
The midpoint between the vectors $\bfa$ and $\bfc$ is $\frac{\bfa + \bfc}{2}$.
Similarly, the other midpoint depicted is $\frac{\bfb + \bfc}{2}$:
We are trying to compare the line segements from $\bfa$ to $\bfb$ and from $\frac{\bfa+\bfc}{2}$ and $\frac{\bfb+\bfc}{2}$. We now write both as vectors.
Recall that
Subtraction gives the vector between two points
The vector from $\bfa$ to $\bfb$ is $\bfb - \bfa$.
Similarly, the vector from $\frac{\bfa+\bfc}{2}$ and $\frac{\bfb+\bfc}{2}$ is $$ \frac{\bfb+\bfc}{2} - \frac{\bfa+\bfc}{2} = \frac{\bfb - \bfa}{2}$$
Vector-scalar multiplication scales the length of the vector
This vector is half the length, but in the same direction as $\bfb - \bfa$.
We conclude that the line segment connecting the midpoints of two sides of a triangle is parallel and half the length of the third side.
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