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  1. #1
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    tangents

    . If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.
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  2. #2
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    Quote Originally Posted by ice_syncer View Post
    . If all the sides of a parallelogram toucha circle, show that the parallelogram is a rhombus.
    If by ‘touch’ you mean ‘tangent to’ then note that two tangents from an exterior point to a circle make congruent line segments to the circle. Think about opposite vertices.
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    Quote Originally Posted by Plato View Post
    If by ‘touch’ you mean ‘tangent to’ then note that two tangents from an exterior point to a circle make congruent line segments to the circle. Think about opposite vertices.
    Yes I know thank you, I got it just after I posted it, we need to add the opposite sides and substitute it with the other equal tangents, I wish I knew how to delete posts..
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  4. #4
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    Hello, ice_syncer!

    I think this is what you meant . . .


    If all the sides of a parallelogram touch a circle,
    show that the parallelogram is a rhombus.
    Code:
                   A      E                  B
                    ∆ - * o * - - - - - - - ∆
                   /*           *          /
                H o               *       /
                 *                 *     /
                /                       /
               /*                   *  /
              / *         * O       * /
             /  *                   */
            /                       /
           /     *                 *
          /       *               o F
         /          *           */
        ∆ - - - - - - - * o * - ∆
       D                  G      C
    We have parallelogram ABCD, inscribed circle O,
    . . and points of tangency E,F,G,H.


    Theorem: tangents from an external point to a circle are equal.

    Hence, we have:
    . . {\color{blue}[1]}\;\;AE \:=\:AH
    . . {\color{blue}[2]}\;\;BE \:=\:BF
    . . {\color{blue}[3]}\;\;CG \:=\:CF
    . . {\color{blue}[4]}\;\;DG \:=\:DH

    Add [1] and [4]: . AE + DG \:=\:AH + DH \quad\Rightarrow\quad AE + DG \:=\:AD\;\;{\color{blue}[5]}

    Add [2] and [3]: . BE + CG \:=\:BF + CF \quad\Rightarrow\quad BE + CG \:=\:BC\;\;{\color{blue}[6]}


    Add [5] and [6]: . (AE+BE) + (DG + CG) \:=\:AD + BC

    So we have: . AB + CD \:=\:AD + BC\;\;{\color{blue}[7]}


    Opposite sides of a parallelogram are equal: . AB = CD,\;\;AD = BC

    So [7] becomes: . 2AB \:=\:2AD \quad\Rightarrow\quad AB \:=\:AD


    Since adjacent sides of the parallelogram are equal, all sides are equal.

    Therefore, ABCD is a rhombus.

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