# 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:

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

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

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