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Tangent and Secant: What Is This About?
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Copyright © 1996-2008 Alexander Bogomolny
Tangent and Secant
The applet suggests the following theorem [Honsberger]:
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Assume TA is tangent to a given circle at point A while a secant through T intersect the circle in points B and C. Let the bisector of angle BTA intersect AB in X and AC in Y. Then |
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The problem is very simple. Note that the equality of two line segments we are asked to prove is equivalent to the assertion that triangle XTY is isosceles, which, in turn, is equivalent (by ASA) to saying that the base angles AXY and AYX are equal. These angles are supplementary to angles AXT and CYT, respectively. Consider two triangles: AXT and CYT. They have equal angles at T, so that angles AXT and CYT are equal iff angles XAT and YCT are equal. Extending the sides of the latter, we have to show that angles BAT and ACB are equal. But this is so because one of them (BAT) is formed by a secant AB and a tangent AT, while the other (ACB) is an inscribed angle subtended by the chord AB. Both equal the angular measure of the arc AB (that does not contain C.)
(Angle BAT is said to intercept arc AB, the latter is said to be intercepted by angle BAT. If AD is a diameter of the circle, then AD is perpendicular to AT, whereas triangle ABD is right. Angle BAD is then simultaneously complementary to angles BAT and ADB. However, angles ADB and ACB are inscribed in the same arc and are therefore equal, and so are angles ACB and BAT.)
References
- Honsberger, Mathematical Chestnuts from Around the World, MAA, 2001, p. 143
Copyright © 1996-2008 Alexander Bogomolny
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