# central angle theorem

• Sep 14th 2010, 09:12 PM
MATNTRNG
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.
• Sep 15th 2010, 05:22 AM
Soroban
Hello, MATNTRNG!

Quote:

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

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

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