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?

Question

cindysnifflesuz · Accepted Answer

Step 1Use 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   ∠      A    O    B   which is   ∠      A    O    C   and try showing that the angle between chord AB and tangent at A is the same.Step 2Note, 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          ∘      .

Luciano Webster · Accepted Answer

Step 1As a diagram of the other common method of proof of the Alternate Segment Theorem has not yet been posted, here it is.Step 2AB is the diameter and so   ∠  B  C  A is a right angle. Hence   y  =  π      /    2  −  ∠  B  A  C  =  x  ,, and we just note that the angle subtended by AC is constant in this segment.

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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