Equal Angles in Two Circles: What is this about?
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Copyright © 1996-2012 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:
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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 © 1996-2012 Alexander Bogomolny| 40620096 |
