# Proving parallel lines using points and vectors

#### algebruh

##### New member
Hey, this is a problem given to me by my prof for an assignment, and the TAs at my tutorials haven't been much help. Was wondering where to go with this question.

Also, I'm a uni freshman who isn't used to the whole concept of proofs, and a lot of what my profs say seem to be a slew of symbols and numbers before they even define anything, but I do the textbook readings and can comprehend those fairly easily. Was anyone on here's transition from high school math to university math a massive jump?

#### skeeter

##### Well-known member
MHB Math Helper
I'm going to assume you were exposed to the triangle proportionality theorem in a HS geometry course. To complete your proof, we require the use of its converse ... if you are not familiar or need a refresher, visit the attached link below.

https://sites.google.com/site/mrros...verse-of-the-triangle-proportionality-theorem

For your problem, if we can show $\dfrac{OA}{AC} = \dfrac{OP}{PR}$, then by the converse of the triangle proportionality theorem $AP \parallel CR$.

Starting with $\Delta OCQ$, you are given $BP \parallel CQ$. Using the triangle proportionality theorem ...

$\dfrac{OB}{BC} = \dfrac{OP}{PQ} \implies \dfrac{OA+AB}{BC} = \dfrac{OP}{PQ} \implies \color{red}{(OA)(PQ)+(AB)(PQ)=(BC)(OP)}$

Same drill with $\Delta OBR$ ...

$\dfrac{OA}{AB} = \dfrac{OQ}{QR} \implies \dfrac{OA}{AB} = \dfrac{OP+PQ}{QR} \implies \color{red}{(OP)(AB)+(PQ)(AB)=(OA)(QR)}$

subtracting the second red equation from the first yields ...

$\color{red}(OA)(PQ)-(OP)(AB)=(BC)(OP)-(OA)(QR)$

rearranging ...

$\color{red} (OA)(PQ)+(OA)(QR) = (BC)(OP)+(OP)(AB)$

factoring both sides ...

$\color{red} (OA)(PQ+QR) = (OP)(AB+BC) \implies (OA)(PR) = (OP)(AC) \implies \dfrac{OA}{AC}= \dfrac{OP}{PR} \implies AP \parallel CR$