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Math Help - Parallelogram and circle

  1. #1
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    Parallelogram and circle

    There is a parallelogram. Its vertexes A, D and B are tangent to the circle (triangle ADB is inscribed in the circle). Circle intersects parallelogram at points K and L. AN is diameter of the circle. Prove that |KN| = |KL|.
    Parallelogram and circle-figure.png
    Last edited by aleschio; September 23rd 2012 at 11:45 AM.
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  2. #2
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    Re: Parallelogram and circle

    SK,SN,and SL are radii and therefore equal. The two triangles are congruent. Congruent chords imply congruent arcs
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  3. #3
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    Re: Parallelogram and circle

    Quote Originally Posted by aleschio View Post
    There is a parallelogram. Its vertexes A, D and B are tangent to the circle

    You mean A, D, and B lie on the circle. Points cannot be 'tangent'.

    (triangle ADB is inscribed in the circle). Circle intersects parallelogram at points K and L. AN is diameter of the circle. Prove that |KN| = |KL|.
    Click image for larger version. 

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  4. #4
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    Re: Parallelogram and circle

    It's all about the angles.

    Repeated use of isosoles triangles gives:

    m\angle LSD + 2m\angle LDS = 180, and m\angle ASD + 2m\angle SAD = 180

    m\angle ASB + 2m\angle SAB = 180, and m\angle BSK + 2m\angle SBK = 180

    Thus:

    (m\angle ASD + m\angle LSD) + 2(m\angle LDS + m\angle SAD) = 360

    (m\angle ASB + m\angle BSK) + 2(m\angle SAB + m\angle SBK) = 360

    From parallelogram m \angle LDA = m\angle ABK, so

    m\angle LDS + m\angle SDA = m\angle ABS + m\angle SBK

    But via isosoles, m\angle SDA = m\angle SAD, and m\angle ABS = m\angle SAB.

    Thus

    m\angle ASD + m\angle LSD

    = 360 - 2(m\angle LDS + m\angle SAD)

    = 360 - 2(m\angle LDS + m\angle SDA)

    = 360 - 2(m\angle ABS + m\angle SBK)

    = 360 - 2(m\angle SAB + m\angle SBK)

    = m\angle ASB + m\angle BSK.

    Now, using that \overline{AN} is a diameter, get

    m\angle ASD + m\angle LSD+ m\angle NSL = 180

    m\angle ASB + m\angle BSK+ m\angle NSK = 180

    Thus

    m\angle NSL = 180 - (m\angle ASD + m\angle LSD)

    = 180 - (m\angle ASB + m\angle BSK)

    = m\angle NSK.

    But from m\angle NSL = m\angle NSK and Side-Angle-Side, get

    \triangle KSN \cong \triangle LSN, and so m\overline{NL} = m\overline{NK}.
    Last edited by johnsomeone; September 26th 2012 at 07:24 PM.
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