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Common Chord and a Tangent: What is this about?
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The applet may suggest the following statement:
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Let there be two circles (O) and (Q) -- in notations that show their centers. Assume the circles intersect and AB is their common chord. Let BM be a piece of tangent to (O) inside (Q). If Q lies on (O), then |
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Proof
Assume Q is on (O). Join QA, QB, and QM. All three are radii of (Q) and hence are equal. Triangles AQB and BQM are isosceles. In addition, their base angles coincide. Indeed, 
ABM between tangent and chord AB cuts off arc AQB and equals half the angular measure of the latter. Inscribed ABQ that is subtended by arc AQ which is one half of arc AQB, is ½ABM. It follows that BM is the bisector of angle ABM and ABQ = MBQ.
References
- R. Nelsen, Proofs Without Words, MAA, 1993, p. 18
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