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Explanation

The applet may suggest the following statement:

 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 AB = BM.

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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 ½∠ABM. It follows that BM is the bisector of angle ABM and ∠ABQ = ∠MBQ. ΔAQB = ΔBQM and AB = BM.

References

1. R. Nelsen, Proofs Without Words, MAA, 1993, p. 18