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Math Help - central angle theorem

  1. #1
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    central angle theorem

    Use the angles sum of a triangle theorem and the linear pair theorem to prove the following result:

    If triangle ABC is a right triangle with right angle at C and O is the midpoint of AB, then angle BOC = 2 * angle BAC.
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  2. #2
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    Hello, MATNTRNG!

    \text{Use the angles sum of a triangle theorem and the linear pair theorem}
    \text{to prove the following result:}

    \text{If }\Delta ABC\text{ is a right triangle with }C = 90^o\text{ and }O\text{ is the midpoint of }AB,
    \text{then: }\:\angle BOC \:=\: 2\!\cdot\!\angle BAC

    I have my own proof of this . . .
    Code:
        A o
          * *   r
          * @ *
          *     *   O
          *       o
          *     * 2@*   r
          * @ * r     *
          * *           *
        C o   *   *   *   o B

    Right triangle ABC can be inscribed in a semicircle.
    . . O is the center of the circle
    . . and OA, OB, OC are radii \,r of the circle.

    Hence, \Delta AOC is isosceles and \angle OAC = \angle ACO = \theta

    \angle BOC is an exterior angle of \Delta AOC.
    . . Hence: . \angle BOC \:=\:\angle OAC + \angle ACO \;=\;\theta + \theta \;=\;2\theta

    Therefore: . \angle BOC \;=\;2\!\cdot\!\angle OAC \;=\;2\!\cdot\!\angle BAC
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