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Explanation

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.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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.