# Tangent-chord theorem

• Apr 9th 2010, 09:48 PM
lpbug
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.
• Apr 10th 2010, 12:29 AM
sa-ri-ga-ma
• Apr 10th 2010, 07:51 AM
Soroban
Hello, lpbug!

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

Quote:

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)
$

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