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Math Help - Prove Points are Concyclic.

  1. #1
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    Prove Points are Concyclic.

    Hey Everyone,

    I have a question here that asks:

    Through a point on the diagonal of a square, lines PR, QS are drawn parallel to the sides, P, Q, R, S being on the sides. Prove that these four points are concyclic.

    Any help would be greatly appreciated. Thanks everyone.
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  2. #2
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    Let O be the point of intersection of the lines PR and QS (which is on the diagonal of the square). Then let the measure of OPS = a and add the measures of the following angles:

    OPS + OPQ = SPQ = a + 45

    OQP + OQR = PQR = 45 + a

    ORQ + ORS = QRS = (90 - a) + 45 = 135 - a

    OSR + OSP = RSP = 45 + (90 - a) = 135 - a

    Then SPQ + QRS = 180 and PQR + RSP = 180, proving that PQRS is a cyclic quadrilateral (i.e., P, Q, R, S are concyclic).
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  3. #3
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    Hello, GreenDay14!

    I have an analytic proof . . .


    Through a point on the diagonal of a square,
    lines PR, QS are drawn parallel to the sides; P, Q, R, S being on the sides.
    Prove that these four points are concyclic.
    Code:
          |             Q
     (0,a)|           (b,a)    (a,a)
          ͸ - - - - - - o - - ͸
          |             |   * |
      P   |             | *   |   R
    (0,b) o - - - - - - * - - o (a,b)
          |           * |     |
          |         o   |     |
          |       * C   |     |
          |     *       |     |
          |   *         |     |
          | *           |     |
    - - - ͸ - - - - - - o - - ͸ - - -
      (0,0)           (b,0)    (a,0) 
                        S

    Let a = side of the square.

    Let (b,b) be the point on the diagonal.

    Then we have: . P(0,b),\;Q(b,a),\;R(a,b),\;S(b,0)

    Let C\left(\tfrac{a}{2},\:\tfrac{a}{2}\right) be the center of the square.


    Find the distances: . CP,\:CQ,\:CR,\:CS

    . . CP^2 \:=\:\left(\tfrac{a}{2}-0\right)^2 + \left(\tfrac{a}{2}-b\right)^2

    . . CQ^2 \:=\:\left(\tfrac{a}{2}-b\right)^2 + \left(a - \tfrac{a}{2}\right)^2

    . . CR^2 \:=\:\left(\tfrac{a}{2}-a\right)^2 + \left(\tfrac{a}{2}-b\right)^2

    . . CS^2 \:=\:\left(\tfrac{a}{2}-b)\right)^2 + \left(\tfrac{a}{2} - 0\right)^2


    We find that they are all equal to: . \tfrac{1}{2}a^2 - ab + b^2


    Hence, there is a point C equidistant from P,Q,R,S.
    . . C is the circumcenter of quadrilateral PQRS.

    Therefore, P,Q,R,S are concyclic.

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