### Equal Angles in Two Circles: What is this about?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander BogomolnyThe 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 |

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

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Copyright © 1996-2018 Alexander Bogomolny64499628 |