Proving parallel lines using points and vectors

In summary, the speaker is struggling with a math problem given by their professor and is not receiving much help from their TAs. They also mention that they find proofs difficult to understand and are unsure of how to approach this question. They then provide a link to the triangle proportionality theorem and explain how it can be used to solve the problem. Finally, they provide a step-by-step solution to the problem, showing how to use the theorem to prove that $AP \parallel CR$.
  • #1
algebruh
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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.

View attachment 9264

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?
 

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  • #2
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/mrrosessite/geometry/geometry-objectives/-g-srt-similarity-right-triangles-and-trigonometry/srt4/converse-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 ...

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

1. How do you prove parallel lines using points?

Parallel lines can be proven using the slope formula. If two lines have the same slope, they are parallel. To find the slope, you can choose any two points on each line and use the formula (y2-y1)/(x2-x1). If the slopes are equal, the lines are parallel.

2. Can parallel lines be proven using vectors?

Yes, parallel lines can also be proven using vectors. If the cross product of two vectors is equal to zero, the lines they represent are parallel. This is because the cross product of two parallel vectors is always equal to zero.

3. What is the difference between proving parallel lines using points and vectors?

The main difference is the method used. Proving parallel lines using points involves finding the slope of each line, while proving parallel lines using vectors involves finding the cross product of two vectors. Additionally, proving parallel lines using points requires knowledge of the slope formula, while proving parallel lines using vectors requires knowledge of vector operations.

4. Are there any other methods for proving parallel lines?

Yes, there are other methods for proving parallel lines. One method is to use the alternate interior angles theorem, which states that if two parallel lines are cut by a transversal, the alternate interior angles are congruent. Another method is to use the parallel lines postulate, which states that if two lines are cut by a transversal and the alternate interior angles are congruent, the lines are parallel.

5. Can parallel lines be proven using only one point?

No, parallel lines cannot be proven using only one point. To prove parallel lines, you need at least two points on each line to find the slope or the cross product of two vectors. Having only one point does not provide enough information to determine the relationship between two lines.

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