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Equal Angles in Two Circles: What is this about?
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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 = BTPATP = 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:
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PST = BPT.
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Now, in ΔBPT,
| (1) |
PBT + BPT + BTP = 180o.
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On the other hand, since quadrilateral ABTS is cyclic,
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AST + ABT = 180o,
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or, which is the same
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ASP + PST + ABT = 180o,
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Comparing (1) and (2) we get
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ASP = BTP.
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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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