Geometric Proofs Regarding Vectors

This page is intended to be a part of the Calculus hub. More in depth math on vectors and matrices can be found on the Linear Algebra hub.

# Geometric Proofs Regarding Vectors

We will now look at some important uses of vectors with regards to some important geometric arguments.

## Example 1

Prove that if $ABCD$ define a quadrilateral such that $P$ is the midpoint of of $AB$, $Q$ is the midpoint of $BC$, $R$ is the midpoint of $CD$, and $S$ is the midpoint of $DA$, then $PQRS$ is a parallelogram.

To solve this problem we will use the following picture of an arbitrary quadrilateral that satisfies the conditions in the problem.

To show that $PQRS$ is a parallelogram, we must show that (1) $\vec{PS} = \vec{QR}$ and (2) $\vec{PQ} = \vec{SR}$ are parallel and have the same length.

First we will show (1). Note that $\vec{PA} = \frac{1}{2} \vec{BA}$ and $\vec{AS} = \frac{1}{2} \vec{AD}$. Therefore $\vec{PQ} = \vec{PA} + \vec{AS} = \frac{1}{2} \vec{BA} + \frac{1}{2} \vec{AD} = \frac{\vec{BA} + \vec{AD}}{2} = \frac{BD}{2}$.

Now also, note that $\vec{QC} = \frac{1}{2} \vec{BC}$ and $\vec{CR} = \frac{1}{2} \vec{CD}$. Therefore $\vec{QR} = \vec{QC} + \vec{CR} = \frac{1}{2} \vec{BC} + \frac{1}{2} \vec{CD} = \frac{\vec{BC} + \vec{CD}}{2} = \frac{\vec{BD}}{2}$.

Therefore we have that $\vec{PS} = \vec{QR}$.

Now we will show (2). Note that $\vec{PB} = \frac{1}{2} \vec{AB}$ and $\vec{BQ} = \frac{1}{2} \vec{BC}$. Therefore $\vec{PQ} = \vec{PB} + \vec{BQ} = \frac{1}{2} \vec{AB} + \frac{1}{2} \vec{BC} = \frac{\vec{AB} + \vec{BC}}{2} = \frac{\vec{AC}}{2}$.

Now also, note that $\vec{SD} = \frac{1}{2} \vec{AD}$ and $\vec{DR} = \frac{1}{2} \vec{DC}$. Therefore $\vec{SR} = \vec{SD} + \vec{DR} = \frac{1}{2} \vec{AD} + \frac{1}{2} \vec{DC} = \frac{\vec{AD} + \vec{DC}}{2} = \frac{\vec{AD}}{2}$.

Therefore we have that $\vec{PQ} = \vec{SR}$.

We therefore conclude that since $PQRS$ forms a quadrilateral such that the opposite sides are equal, then $PQSR$ must be a parallelogram.

## Example 2

Prove that if $ABCD$ form a parallelogram, then the diagonals bisect each other.

To solve this problem we will use the following picture to prove this statement.

To show that the diagonals of $ABCD$ bisect each other we will show that $\vec{PT} = \vec{TR}$ and that $\vec{ST} = \vec{TQ}$.

First notice that $\vec{PA} = \frac{1}{2} \vec{BA}$, $\vec{AS} = \frac{1}{2} \vec{AD}$ and $\vec{ST} = \frac{1}{2} \vec{SQ}$. Therefore $\vec{PT} = \vec{PA} + \vec{AS} + \vec{ST} = \frac{1}{2} \vec{BA} + \frac{1}{2} \vec{AD} + \frac{1}{2} \vec{SQ} = \frac{\vec{BA} + \vec{AD} + \vec{SQ}}{2} = \frac{\vec{BD} + \vec{SQ}}{2}$.

Now notice that $\vec{TS} = \frac{1}{2} \vec{QS}$, $\vec{SD} = \frac{1}{2} \vec{AD}$ and $\vec{DR} = \frac{1}{2} \vec{DC}$. Therefore $\vec{TR} = \vec{TS} + \vec{SD} + \vec{DR} = \frac{1}{2} \vec{QS} + \frac{1}{2} \vec{AD} + \frac{1}{2} \vec{DC} = \frac{\vec{QS} + \vec{AD} + \vec{DC}}{2} = \frac{\vec{QS} + \vec{AC}}{2}$