How to prove the tangent secant theorem

Question

I was reading the the following theorem: Let $A,B$ be two points on the circumference of a circle. Let $C$ be a point outside the circle. Then $\angle BAC=\frac{1}{2}\widehat{AB}$. Is there some elementary way to prove it? Here is a picture with different points: http://jwilson.coe.uga.edu/emt668/EMAT6680.2003.fall/Nichols/6690/Webpage/Day%209.htm, the part "is formed by tangent and secant".

Answer 1

I think you have understand the problem incorrectly. This is the case only when the segment $\overline{AC}$ is tangent to the circle. In this case we have $\measuredangle{BAC}=\frac{1}{2}\widetilde{AB}$, in which $\widetilde{AB}$, denotes the arc $AB$, and its proof is completely straightforward. I you draw the diameter passing from $A$, intersects the other side of the circle in $A'$.

$\measuredangle{BAC}=\frac{\pi}{2}-\measuredangle{BAA'}=\frac{\pi}{2}-\frac{1}{2}\widetilde{A'B}=\frac{\pi-\widetilde{A'B}}{2}=\frac{\widetilde{A'A}-\widetilde{A'B}}{2}=\frac{\widetilde{BA}}{2}$