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    #1
    I can't figure this one out...(I may not have drawn it exactly how it was drawn in the book)


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    #2
    Quote Originally Posted by woof123 View Post
    I can't figure this one out...(I may not have drawn it exactly how it was drawn in the book)
    angle PRQ = 180 - 2x by supplementary angles.

    Thus angle PQR = 180 - (x) - (180 - 2x) = x. What does this tell you?

    -Dan

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    #3 Thread Author
    OK, I get x, meaning the sides opposite to the x angles are equal.

    Thanks so much

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    #4
    Okay, let's label the unknown angles:

    \begin{tikzpicture}
    \draw[blue,thick] (0,0) -- (12,0);
    \draw[blue,thick] (12,0) -- (8,4);
    \draw[blue,thick] (8,4) -- (0,0);
    \draw[blue,thick] (8,4) -- (6,0);
    \node[left=5pt of {(0,0)}] {\large P};
    \node[above=5pt of {(8,4)}] {\large Q};
    \node[below=5pt of {(6,0)}] {\large R};
    \node[right=5pt of {(12,0)}] {\large S};
    \node[right=15pt of {(0,0)},yshift=5pt] {\large $x$};
    \node[right=5pt of {(6,0)},yshift=7pt] {\large $2x$};
    \node[left=5pt of {(8,4)},yshift=-13pt] {\large $\alpha$};
    \node[below=3pt of {(8,4)}] {\large $\beta$};
    \node[left=10pt of {(12,0)},yshift=7pt] {\large $\delta$};
    \node[above=0pt of {(6,0)},xshift=-3pt] {\large $\gamma$};
    \end{tikzpicture}

    Now, we know the following:

    $ \displaystyle 2x+\gamma=180^{\circ}$

    $ \displaystyle x+\alpha+\gamma=180^{\circ}$

    $ \displaystyle 2x+\beta+\delta=180^{\circ}$

    $ \displaystyle x+\alpha+\beta+\delta=180^{\circ}$

    We need to show that:

    $ \displaystyle x=\alpha$

    Look at the first two equations above, and we see that:

    $ \displaystyle 180^{\circ}-\gamma=2x=x+\alpha$

    What does this imply?

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    #5
    I would like to express my opinion here

    Theorem: The exterior angle formed when a side of a triangle is produced
    is equal to the sum of the two interior opposite angles.

    Now using this theorem,

    $2x=x+\angle PQR$
    $2x-x= \angle PQR$
    $x= \angle PQR$

    Then we may recall the theorem,

    Theorem (Converse of the theorem on isosceles triangles):
    The sides opposite equal angles of a triangle are equal

    Now since we have proved that $ \angle PQR =\angle QPR$ what can be said about the sides PQ and PR ?

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