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Thread: 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!

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

    $\displaystyle \text{If }\Delta ABC\text{ is a right triangle with }C = 90^o\text{ and }O\text{ is the midpoint of }AB,$
    $\displaystyle \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 $\displaystyle ABC$ can be inscribed in a semicircle.
    . . $\displaystyle O$ is the center of the circle
    . . and $\displaystyle OA, OB, OC$ are radii $\displaystyle \,r$ of the circle.

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

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

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