Angle between chord and tangent I am not getting this theorem:

Angle between chord AB and tangent at A is the same as subtended by segment AB at any point on the circumference.

How to prove this theorem?
 A: Use the fact that if $O$ is the center of circle, and $C$ is foot of the perpendicular from $O$ to $AB$, then the angle subtended at any point on the circumference is half $\angle{AOB}$ which is $\angle{AOC}$ and try showing that the angle between chord $AB$ and tangent at $A$ is the same.

Note, you do have to be careful here, there are two different angles subtended by points on the circumference, corresponding to the two different arcs which $AB$ forms. In fact those two angles add up to $180^{\circ}$.
A: As a diagram of the other common method of proof of the Alternate Segment Theorem has not yet been posted, here it is.

AB is the diameter and so $ \angle BCA $ is a right angle. Hence $y = \pi/2 - \angle BAC = x,$
and we just note that the angle subtended by $AC$ is constant in this segment.
A: It is well known that the angle subtended by segment $AB$ at any point on the circumference is constant on either of the arcs $AB$. Since the sum of those angles is $\pi$, it suffices to prove this for the point $C$ that is opposite to point $A$. Then it is well known that the angle $ABC$ is a right angle and also the segment $CA$ meets the tangent at $A$ in a right angle. Therefore $CAB = \frac{\pi}{2} - BCA$ and your angle is then just $\frac{\pi}{2} - CAB = BCA$.
A: The angle between chord AC and tangent is half of ∠AOC.
To see this, let's expand on Derek Jennings answer @DerekJennings ,
https://i.stack.imgur.com/1eFkR.png
Derek Jennings diagram
Let point M denote the midpoint of chord AC. Note that a line OM is perpendicular to AC. Therefore OM is parallel to alternate chord BC. 
Considering the diameter AB as a traversal of parallel lines OM and BC, identical angles are created at both intersections.
Since by definition OM bisects ∠AOC.
