Pole and Polar with Respect to a Triangle
What is this about?
A Mathematical Droodle
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander BogomolnyPole and Polar with respect to a Triangle
The applet illustrates a variant of Desargues' theorem.
In ΔABC cevians AA', BB', CC' meet in point P. The lines AB, A'B' meet in C_{0}, AC, A'C' meet in B_{0} and BC and B'C' meet in A_{0}. Then the points A_{0}, B_{0}, C_{0} are collinear.
The converse is also true.
The fact that the given cevians are concurrent, tells us that the triangle ABC and A'B'C' are perspective from a point (P in our notations. A'B'C' is a cevian triangle of P.) By Desargues' theorem, the triangles are also perspective from a line. In other words, the points A_{0}, B_{0}, C_{0} of intersection of the corresponding sides of the two triangles are collinear. The line they lie on is called the polar of point P with respect to ΔABC. Point P is known as the pole of that line with respect to ΔABC. Sometimes they are referred to as trilinear pole and polar with respect to the triangle.
(This is just a generalization of the statement concerning Gergonne point and Gergonne line.)
It was noted by Maj/Cpt/Lt Pestich that the above statement highlights the relationship between the Ceva and Menelaus theorems. From the construction, the pairs A_{0} and A', B_{0} and B', C_{0} and C' are harmonic conjugate with respect to (B, C), (C, A), (A, B), respectively. Formally, using the cross-ratio (and signed segments)
(UVWX) = WU/WV : XU/XV,
we directly obtain
(ABC'C_{0}) = C'A/C'B : C_{0}A/C_{0}B = -1,
(BCA'A_{0}) = A'B/A'C : A_{0}B/A_{0}C = -1,
(CAC'B_{B}) = B'C/B'A : B_{0}C/B_{0}A = -1.
In other words,
AC'/C'B = -AC_{0}/C_{0}B,
BA'/A'C = -BA_{0}/A_{0}C,
CB'/B'A = -CB_{0}/B_{0}A.
These we multiply:
(1) | AC'/C'B · BA'/A'C · CB'/B'A = - AC_{0}/C_{0}B · BA_{0}/A_{0}C · CB_{0}/B_{0}A. |
So that if one of the products equals 1, the other equals -1 such that the two statements below are equivalent:
AC'/C'B · BA'/A'C · CB'/B'A = 1 and
AC_{0}/C_{0}B · BA_{0}/A_{0}C · CB_{0}/B_{0}A = -1.
Thus, if three points on the sides of a triangle satisfy Ceva's concurrency condition, their harmonic conjugates with respects to the vertices of the triangle satisfy Menelaus' collinearity condition, and vice versa. The two theorems are indeed equivalent.
In terms of cross-ratios (1) can be rewritten as
(ABC'C_{0})·(BCA'A_{0})·(CAB'B_{0}) = -1.
Poles and Polars
- Poles and Polars
- 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
- La Hire's Theorem in Ellipse
- Nobbs' Points, Gergonne Line
- Polar Circle
- Pole and Polar with Respect to a Triangle
- Poles, Polars and Quadrilaterals
- Straight Edge Only Construction of Polar
- Tangents and Diagonals in Cyclic Quadrilateral
- Secant, Tangents and Orthogonality
- Poles, Polars and Orthogonal Circles
- Seven Problems in Equilateral Triangle, Solution to Problem 1
Desargues' Theorem
- Desargues' Theorem
- 2N-Wing Butterfly Problem
- Cevian Triangle
- Do You Speak Mathematics?
- Desargues in the Bride's Chair (with Pythagoras)
- Menelaus From Ceva
- Monge from Desargues
- Monge via Desargues
- Nobbs' Points, Gergonne Line
- Soddy Circles and David Eppstein's Centers
- Pascal Lines: Steiner and Kirkman Theorems II
- Pole and Polar with Respect to a Triangle
- Desargues' Hexagon
- The Lepidoptera of the Circles
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny64641849 |