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Math Help - Circle Geometry: Proving a line passes through centre of circle

  1. #1
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    Circle Geometry: Proving a line passes through centre of circle

    AB and CD are two parallel chords of a circle. CA and DB meet, when produced, at X. AD and BC intersect at Y

    There are three parts to this question, two of which are already solved and i will post results:

    a)YA= YB
    b)XA= XB

    How must i prove that XY will pass through center?

    And just generally, how is this achieved without using co-ordinate geometry, i.e. proving a certain line passes through centre.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Hint:

    prove XY radius .
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    Hmm... my X is outside the circle so it definately does not equate to radius. Do you have any other ideas?
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    AB is parallel to CD.

    So \angle{BAD} + \angle{DCA} = 180^o

    ABDC is cyclic quadrilateral.

    So \angle{BAC} + \angle{BDC} = 180^o

    Hence \angle{ACD} = \angle{BDC}

    \angle{ACB} and \angle{ADB} on the chord AB are equal.

    Hence \angle{yCD} = \angle{yDC}. So yC = yD.

    If O is the center of the circle, then OC = OD. It is possible only if O and y lie on the perpendicular bisector of CD.

    Hence xy passes from the center of the circle.
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    So the reasnoning here is basically that; since YC = YD, this can only at one point in the circle. Therefore Y is O?
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    Quote Originally Posted by Lukybear View Post
    So the reasnoning here is basically that; since YC = YD, this can only at one point in the circle. Therefore Y is O?
    Y is not O. Here OC = OD = radius of the circle. And YC = YD. It is possible when Y lies on OX.
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  7. #7
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    [ a wordy answer- interested in comments

    two parallel chords subtend equal arcs therefore ABCD is a regular trapezoid. Its diagonals are equal and meetalong the perpendicular bisector of AB,CD and the center of circle.
    angle ACD = angle BCD(equal arcs) triangle XCD is isosocles therefore X lies along the same perpendicular bisector


    bjh
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