Equal Angles in Two Circles II: What is this about?
A Mathematical Droodle
| |
|
Explanation
|Activities|
|Contact|
|Front page|
|Contents|
|Store|
|Geometry|
Copyright © 1996-2012 Alexander Bogomolny
The applet presents a further generalization of a problem of an accidental angle bisector. The latest generalization is due to Nathan Bowler who has observed that a tangent to the inner circle can be replaced by a chord thus making the problem perfectly symmetric with respect to the two circles.
| |
Given two circles that cross at points S and T and a chord AB of one that crosses the other in points P and Q as shown. Then ∠ASP = ∠BTQ (and also ∠ATP = ∠BSQ).
|
The proof makes use of the fundamental property of inscribed quadrilaterals. Note that there are two of them: ASTB and PSTQ.
| |
|
We also apply the Exterior Angle Theorem:
| (1) |
| ∠ASP | = ∠AST - ∠PST |
| | = (π - ∠ABT) - (π - ∠PQT) |
| | = ∠BTQ. |
|
Now, angles ASB and ATB are equal as inscribed in one of the circles while angles PSQ and PTQ are equal as inscribed in the other. Subtracting twice equal angles from equal angles we get an additional identity:
|Activities|
|Contact|
|Front page|
|Contents|
|Store|
|Geometry|
Copyright © 1996-2012 Alexander Bogomolny
| 
