Hello, lpbug!

Why don't you show us your proof?

Maybe we can find a *different* one?

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 $\displaystyle O.$

The chord is $\displaystyle AC$, the tangent is $\displaystyle AB.$

Let $\displaystyle \theta = \angle COA,\;\alpha = \angle CAB$

$\displaystyle \theta$ is a central angle, hence: .$\displaystyle \text{arc}(AC) = \theta$

$\displaystyle \Delta COA$ is isosceles.

. . Hence: .$\displaystyle \angle OAC \:=\:\frac{180^o - \theta}{2} \:=\:90^o - \tfrac{1}{2}\theta$

$\displaystyle \angle OAB = 90^o$, so: .$\displaystyle \alpha \:=\:90^o - \angle OAC \:=\:90^o - \left(90^o - \tfrac{1}{2}\theta\right)

$

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