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Thread: Intersecting lines

  1. #1
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    Intersecting lines

    In the figure below, transversal $\displaystyle r$ crosses both $\displaystyle p$ and $\displaystyle q$, and angle $\displaystyle a$ and $\displaystyle b$ are measures of the indicated angles, both between $\displaystyle 0$ and $\displaystyle 180$. Lines $\displaystyle p$ and $\displaystyle q$ will cross somewhere to the left of transversal $\displaystyle r$ (that is, on the side opposite the indicated angles).
    Question: Why is $\displaystyle a + b > 180$ a true relationship between $\displaystyle a$ and $\displaystyle b$ for all possible positions of transversal $\displaystyle r$?
    Attached Thumbnails Attached Thumbnails Intersecting lines-lines.jpg  
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  2. #2
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    Hello, dannyc!

    $\displaystyle \text}In the figure below, transversal }r\text{ crosses both }p\text{ and }q,$
    $\displaystyle \text{and angles }a\text{ and }b\text{ are measures of the indicated angles,}$
    $\displaystyle \text{both between }0^o\text{ and }180^o.$

    $\displaystyle \text{Lines }p\text{ and }q\text{ intersect somewhere to the left of transversal }r.$

    $\displaystyle \text{Why is }a + b \,>\, 180^o\text{ true for all possible positions of transversal }r\,?$

    Code:
                                       r
                                      /       p 
                                   K /    *
                                    o
                             *     / a
                      *           /
               *                 /
      J o                       /
               *               /
                      *       / b
                             o
                            / L    *
                                         *
                                             q

    Let: .$\displaystyle \begin{Bmatrix} p \cap q &=& J \\
    p \cap r &=& K \\ q \cap r &=& L \end{Bmatrix}$

    We have: .$\displaystyle \begin{Bmatrix}\angle JKL \,=\,180^o - a \\ \angle JLK \,=\,180^o - a \end{Bmatrix}$

    Hence: .$\displaystyle \angle J \;=\; 180^o - (180^o - a) - (180^o - b) \;=\;(a+b) - 180^o$


    Since $\displaystyle \angle J$ is a positive angle: .$\displaystyle (a+b)-180^o \:>\:0^o$

    Therefore: .$\displaystyle a + b \:>\:180^o$

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Since $\displaystyle \angle J$ is a positive angle: .$\displaystyle (a+b)-180^o \:>\:0^o$

    Thanks Soroban! I think I got it now! (Btw, JLK = 180 - b above)
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