Secant, Tangents and Orthogonality:
What Is This About?
A Mathematical Droodle
What if applet does not run? |
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Copyright © 1996-2018 Alexander Bogomolny
Secant, Tangents and Orthogonality
The applet suggests the following theorem:
Point P lies outside a circle with center O. Straight lines L_{1} and L_{2} intersect at P. Line L_{1} touches the circle at point A and L_{2} intersects the circle at points B and C. Tangents to the circle at points B and C intersect at point X. Prove that segments AX and OP are perpendicular.
What if applet does not run? |
Proof 1 (Elementary)
Denote angle AOP by φ, PCO by α, and COP by &beta. Then the law of sines implies
(1) | OP/sinα = CO/sinβ = OP·sinφ/sinβ. |
Consider the projections of points A and X on line OP - points A' and X' (not shown in the applet.) Calculate the lengths of PA' and PX':
(2) | PA' = OP·cos^{2}φ, |
whereas
(3) |
So we see that A' coincides with X'. But the projections of two points on the line OP can coincide if and only if these points lie on the line perpendicular to OP. |
Proof 2
We know that A and X are the poles of L_{1} and L_{2}, respectively. The point of intersection of the two lines (P) lies therefore on the the polars of both A and X. By La Hire's theorem, both A and X lie on the polar of P. But a polar of a point with respect to a circle is perpendicular to the line joining the point with the center of the circle. Q.E.D.
References
- D. Fomin, A. Kirichenko, Leningrad Mathematical Olympiads 1987-1991, MathPro Press, 1994, pp. 45, 159.
Poles and Polars
- Archimedes' Twin Circles and a Brother
- Brianchon's Theorem
- Complete Quadrilateral
- Harmonic Ratio
- Harmonic Ratio in Complex Domain
- Inversion
- Joachimsthal's Notations
- La Hire's Theorem
- La Hire's Theorem, a Variant
- Nobbs' Points, Gergonne Line
- Pole and Polar with respect to a Triangle
- Polar Circle
- Poles, Polars and Quadrilaterals
- Secant, Tangents and Orthogonality
- Straight Edge Only Construction of Polar
- Tangents and Diagonals in Cyclic Quadrilateral
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Copyright © 1996-2018 Alexander Bogomolny68955500