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.

Screen%20Shot%202014-11-23%20at%208.43.05%20PM.png

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.

Screen%20Shot%202014-11-23%20at%209.15.39%20PM.png

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 $\vec{TR} = \vec{TS} + \vec{SD} + \vec{DR} = \frac{1}{2} \vec{QS} + \frac{1}{2} \vec{AD} + \frac{1}{2} \vec{DC} = \frac{\vec{QS} + \vec{AD} + \vec{DC}}{2} = \frac{\vec{QS} + \vec{AC}}{2}$

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