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Math Help - Geometric Proof

  1. #1
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    Geometric Proof

    1. The vertex E of a square EFGH is inside a square ABCD. The vertices F, G and H are outside the square ABCD. The side EF meets the side CD at X and the side EH meets the side AD at Y. If EX = EY, prove that E lies on BD.

    2. Two circles C1 and C2 meet at the points A and B. The tangent to C1 at A meets C2 at P. A point Q inside C1 lies on the circumference of C2. When produced, BQ meets C1 at S and PA produced at T. Prove that AS is parallel to PQ.

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  2. #2
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    Re: Geometric Proof

    Hello, john128!

    1. The vertex E of a square EFGH is inside a square ABCD.
    The vertices F, G and H are outside the square ABCD.
    The side EF meets the side CD at X and the side EH meets the side AD at Y.
    If EX = EY, prove that E lies on BD.

    Code:
                    A               B
                H   o---------------o
                o   |               |
               *   Y*               |
              *     |θ  *           |
             *     Z+ - - - oE      |
            *       |      *:       |
           *        |     * :       |
        G o         |    *θ :       |
              *   D o---*---+-------o C
                  *    *X   W 
                      o
                      F
    Draw EW \perp CD.
    Draw EZ \perp AD.
    Let \angle EXW \,=\,\theta \,=\,\angle EYZ.

    In \Delta EW\!X\!:\;EW \,=\,EX\sin\theta
    In \Delta EZY\!:\;EZ \,=\,EY\sin\theta

    Since EX = EY, then EW = EZ.

    Hence, E is equidistant from sides AD and CD.
    That is, E is on the bisector of \angle C.

    Therefore, E is on the diagonal BD.
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  4. #4
    MHF Contributor red_dog's Avatar
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    Re: Geometric Proof

    Quote Originally Posted by john128 View Post
    2. Two circles C1 and C2 meet at the points A and B. The tangent to C1 at A meets C2 at P. A point Q inside C1 lies on the circumference of C2. When produced, BQ meets C1 at S and PA produced at T. Prove that AS is parallel to PQ.

    In the circle C_2, \ \widehat{BQP}=\widehat{BAP}=\frac{arc(BP)}{2}.
    But in the circle C_1, \ \widehat{BAP}=\frac{arc(AB)}{2} and \widehat{BSA}=\frac{arc(AB)}{2}.
    Then \widehat{BQP}=\widehat{BSA}\Rightarrow PQ \parallel AS
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