Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 2 of 2
  1. MHB Apprentice

    Status
    Offline
    Join Date
    Sep 2019
    Posts
    1
    Thanks
    0 times
    Thanked
    0 times
    #1
    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?

  2. # ADS
    Circuit advertisement
    Join Date
    Always
    Posts
    Many
     

  3. MHB Craftsman
    MHB Math Helper
    skeeter's Avatar
    Status
    Offline
    Join Date
    Mar 2012
    Location
    North Texas
    Posts
    382
    Thanks
    41 times
    Thanked
    484 times
    Thank/Post
    1.267
    Awards
    MHB Calculus Award (2017)
    #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.





    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 ...

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

Similar Threads

  1. [SOLVED] Distance between parallel lines
    By mathmari in forum Linear and Abstract Algebra
    Replies: 8
    Last Post: July 15th, 2017, 12:14
  2. [SOLVED] s6.803.12.5.9 are lines parallel?
    By karush in forum Calculus
    Replies: 1
    Last Post: December 28th, 2016, 17:24
  3. slope and parallel lines
    By flnursegirl in forum Pre-Algebra and Algebra
    Replies: 5
    Last Post: July 10th, 2016, 17:32
  4. [SOLVED] Circles and parallel Lines
    By wailingkoh in forum Geometry
    Replies: 5
    Last Post: July 26th, 2015, 21:12
  5. Distance between two non parallel lines
    By mathmari in forum Calculus
    Replies: 0
    Last Post: March 18th, 2015, 17:27

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards