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Math Help - Tangent-chord theorem

  1. #1
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    Tangent-chord theorem

    The measure of a tangent-chord angle is half the measure of the intercepted arc inside the angle.

    Can anyone prove this?
    I've been trying to do it for the whole afternoon, help would be much appreciated, thanks.

    =edit=
    figured out the proof, trying to find a new one now.
    Last edited by lpbug; April 9th 2010 at 10:15 PM. Reason: Figured out.
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  2. #2
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    Show your proof.
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  3. #3
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    Hello, lpbug!

    Why don't you show us your proof?
    Maybe we can find a different one?


    The measure of a tangent-chord angle
    is half the measure of the intercepted arc inside the angle.
    Code:
                  * * * 
              *           *
            *               * C
           *                 o
                          * /
          *       O    *   /  *
          *         o     /   *
          *         | θ  /    *
                    |   /
           *        |  /     * θ
            *       | /     *
              *     |/ α  *
          - - - - * o * - - - - B
                    A

    We have a circle with center O.
    The chord is AC, the tangent is AB.

    Let \theta = \angle COA,\;\alpha = \angle CAB

    \theta is a central angle, hence: . \text{arc}(AC) = \theta


    \Delta COA is isosceles.
    . . Hence: . \angle OAC \:=\:\frac{180^o - \theta}{2} \:=\:90^o - \tfrac{1}{2}\theta


    \angle OAB = 90^o, so: . \alpha \:=\:90^o - \angle OAC \:=\:90^o - \left(90^o - \tfrac{1}{2}\theta\right)<br />

    . . Therefore: . \alpha \;=\;\tfrac{1}{2}\theta . . . . Q.E.D.

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