Equal Angles in Two Circles: What is this about?
A Mathematical Droodle

 

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Explanation

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Copyright © 1996-2017 Alexander Bogomolny

The applet presents a generalization of a recent problem of an accidental angle bisector. (There is a further generalization of this result.) In the present problem the two equal halves of an angle split up but stay equal:

  Given two circles that cross at points S and T and a point P on one but inside the other. Let AB be a chord in the latter touching the former at P. Then ∠ASP = ∠BTP (and also ∠ATP = ∠BSP).

The proof is even simpler than that of the original problem.

 

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The inscribed ∠PST is subtended by chord PT so that it equals the angle formed by the chord with the tangent at its end point P:

  ∠PST = ∠BPT.

Now, in ΔBPT,

(1) ∠PBT + ∠BPT + ∠BTP = 180°.

On the other hand, since quadrilateral ABTS is cyclic,

  ∠AST + ∠ABT = 180°,

or, which is the same

(2) ∠ASP + ∠PST + ∠ABT = 180°,

Comparing (1) and (2) we get

  ∠ASP = ∠BTP.

The equality of angles ATP and BSP from this and the fact that quadrilateral ABTS is cyclic such that ∠ASB = ∠ATB.

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Copyright © 1996-2017 Alexander Bogomolny

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