# Concurrence Not from School Geometry

### Sohail Farhangi

17 February, 2013

Let the incircle \(T\), of \(\Delta ABC\) be tangent to the sides at \(A'\), \(B'\), and \(C'\) as shown below. For a point \(P\) in the plane, let \(A'P\), \(B'P\), and \(C'P\) intersect \(T\) at \(A_1\), \(B_1\), and \(C_1\) respectively. Then \(AA_1\), \(BB_1\), and \(CC_1\) concur at a point \(P'\).

The applet below serves a dynamic illustration:

### Proof

James Tao recommended to me a proof using the trigonometric form of Ceva's Theorem. First we will use the extended Law of Sines on \(\Delta AA_{1}C'\) and \(\Delta A_{1}C'A'\) to obtain the following equations where \(r\) is the radius of \(T\).

(1) | \(\displaystyle \frac{C'A_1}{\mbox{sin}(\angle C'AA_{1})}=\frac{AA_1}{\mbox{sin}(\angle AC'A_{1})} \) |

(2) | \(\displaystyle \frac{C'A_1}{\mbox{sin}(\angle C'A'A_{1})}=2r. \) |

Dividing (2) by (1) gives us the following equation:

(3) | \(\displaystyle \frac{\mbox{sin}(\angle C'AA_{1})}{\mbox{sin}(\angle C'A'A_{1})}=\frac{2r\mbox{sin}(\angle AC'A_{1})}{AA_{1}}. \) |

A similar analysis for \(\Delta AB'A_1\) and \(\Delta A_{1}1B'A'\) will give us the next equation:

(4) | \(\displaystyle \frac{\mbox{sin}(\angle B'AA_{1})}{\mbox{sin}(\angle B'A'A_{1})}=\frac{2r\mbox{sin}(\angle AB'A_{1})}{AA_{1}}. \) |

Now dividing (3) by (4) yields another equation:

(5) | \(\displaystyle \frac{\mbox{sin}(\angle C'AA_{1})\mbox{sin}(\angle B'A'A_{1})}{\mbox{sin}(\angle C'A'A_{1})\mbox{sin}(\angle B'AA_{1})}=\frac{\mbox{sin}(\angle AC'A_{1})}{\mbox{sin}(\angle AB'A_{1})}. \) |

Now note that \(\angle AB'A_{1}=\angle B'A'A_1\) since \(AB'\) is tangent to \(T\), and similarly we have \angle \(AC'A_{1} = \angle C'A'A_1\), which gives us

(6) | \(\displaystyle \frac{\mbox{sin}(\angle C'AA_{1})}{\mbox{sin}(\angle B'AA_{1})}=\bigg(\frac{\mbox{sin}(\angle C'A'A_{1})}{\mbox{sin}(\angle B'A'A_{1})}\bigg)^{2}. \) |

Finally, we are ready to apply the trigonometric form of (the inverse of) Ceva's theorem:

(7) | \(\displaystyle \frac{\mbox{sin}(\angle C'AA_{1})\mbox{sin}(\angle B'CC_{1})\mbox{sin}(\angle A'BB_{1})}{\mbox{sin}(\angle A_{1}AB')\mbox{sin}(\angle C_{1}CA')\mbox{sin}(\angle B_{1}BC')}= \\ \displaystyle\bigg(\frac{\mbox{sin}(\angle PA'C')\mbox{sin}(\angle PC'B')\mbox{sin}(\angle PB'A')}{\mbox{sin}(\angle B'A'P)\mbox{sin}(\angle A'C'P)\mbox{sin}(\angle C'B'P)}\bigg)^{2}=1. \) |

### Menelaus and Ceva

- The Menelaus Theorem
- Menelaus Theorem: proofs ugly and elegant - A. Einstein's view
- Ceva's Theorem
- Ceva in Circumscribed Quadrilateral
- Ceva's Theorem: A Matter of Appreciation
- Ceva and Menelaus Meet on the Roads
- Menelaus From Ceva
- Menelaus and Ceva Theorems
- Ceva and Menelaus Theorems for Angle Bisectors
- Ceva's Theorem: Proof Without Words
- Cevian Cradle
- Cevian Cradle II
- Cevian Nest
- Cevian Triangle
- An Application of Ceva's Theorem
- Trigonometric Form of Ceva's Theorem
- Two Proofs of Menelaus Theorem
- Simultaneous Generalization of the Theorems of Ceva and Menelaus
- Menelaus from 3D
- Terquem's Theorem
- Cross Points in a Polygon
- Two Cevians and Proportions in a Triangle, II
- Concurrence Not from School Geometry
- Two Triangles Inscribed in a Conic - with Elementary Solution
- From One Collinearity to Another
- Concurrence in Right Triangle
|Contact| |Front page| |Content| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

65094996 |