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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?

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up vote 7 down vote accepted

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.

alt text

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 $180^{\circ}$.

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Hey! Can you please put a figure am not getting that abstruse statement :( – Quixotic Dec 3 '10 at 7:53
@Deb: Added a figure. Hope that helps. – Aryabhata Dec 3 '10 at 8:06
It is a shame that one cannot upvote an answer again after it has been edited. :) – J. M. Dec 3 '10 at 8:11
@J.M: Careful...That only removes the previous vote :-) – Aryabhata Dec 3 '10 at 8:14

As a diagram of the other common method of proof of the Alternate Segment Theorem has not yet been posted, here it is.

alt text

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.

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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 $\frac{\pi}{2} - CAB = BCA$.

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