Midpoints of altitudes and concurrent cevians

darij grinberg

Let $ABC$ be a triangle. Let $A_1$ , $B_1$ , $C_1$ be the midpoints of its sides $BC$ , $CA$ , $AB$ , and $A_2$ , $B_2$ , $C_2$ the midpoints of the altitudes from $A$ , $B$ , $C$ . Show that the lines $A_1A_2$ , $B_1B_2$ , and $C_1C_2$ meet at one point.

**Posted:** Mon Jul 05, 2004 5:18 am

_________________
Now the die is cast, the first step taken, a glimmer of hope lights up our lives
Visions of the past, dreams forsaken forming right under our eyes
We are alive...

mecrazywong

Let AD,BE,CF be the altitudes. $B_1,C_1,A_2$ are collinear since they are the midpoints of $AC,AB,AD$ , and thus, we get $\frac {C_1A_2}{A_2B_1} = \frac {BD}{DC}$ by similar triangles. Simiarly, $\frac {A_1B_2}{B_2C_1} = \frac {CE}{EA}$ , $\frac {B_1C_2}{C_2A_1} = \frac {AF}{FC}$ . The result follows by Ceva.

**Posted:** Wed Dec 29, 2004 11:02 pm

jayme · Yang-Mills Theory Offline Joined: 20 Nov 2007 Posts: 681

Dear Fang-jl and Mathlinkers,
let A', B', C' be the feet of the A, B, C-altitude of ABC, K the Lemoine's point of ABC, B1, C1 the midpoints of CA, AB, and U, V the meet points of A'B' with A1C1, B1C1.
According to "Another unlikely concurrence" (http://perso.orange.fr/jl.ayme vol.1), AU and AV go through K.
According to Pappus's theorem, Pappus, KIM) applied to the hexagone AUC1VBA'A, A1, K and A2 are collinear.
….
Sincerely
Jean-Louis

**Posted:** Fri Aug 08, 2008 12:19 am

Quang Tuan Bui

Dear Darij Grinberg,

Let A1A3, B1B3, C1C3 be the altitudes of triangle A1B1C1.
A1A3 = AA2 and A1A3//AA2 therefore AA2A1A3 is parallelogram. Otherwise AC1A1B1 is also parallelogram. It means A2 is reflection of A3 in midpoint of B1C1. Similarly with B2, C2.
A1A3, B1B3, C1C3 are concurrent therefore A1A2, B1B2, C1C2 are concurrent at isotomic conjugate of orthocenter of A1B1C1.

Best regards,
Bui Quang Tuan

**Posted:** Fri Aug 08, 2008 2:07 am

_________________
To discover new interesting: first observe all strange, abnormal but then make them familiar, normal!

Leonhard Euler

**Posted:** Fri Aug 08, 2008 4:08 am

jayme · Yang-Mills Theory Offline Joined: 20 Nov 2007 Posts: 681

Sorry, I made a typos: read BV and not AU
Sincerely
Jean-Louis

**Posted:** Fri Aug 08, 2008 4:32 am

vittasko

It is also true for any arbitrary point $P$ inwardly (or outwardly) to $\bigtriangleup ABC,$ instead of the orthocenter and its cevian triangle $\bigtriangleup A'B'C',$ as a direct application of the Cevian Nests Theorem, because of the midpoints $A_{2},\ B_{2},\ C_{2}$ of $AA',\ BB',\ CC'$ respectively, lie on $B_{1}C_{1},\ A_{1}C_{1},\ A_{1}B_{1}$ and $AA_{2}\cap BB_{2}\cap CC_{2}\equiv P.$

Kostas Vittas.

**Posted:** Fri Aug 08, 2008 5:39 am

vittasko

GENERAL PROBLEM. – A triangle $\bigtriangleup ABC$ is given and let $\bigtriangleup A'B'C'$ be, its cevian triangle, with respect to an arbitrary point $P$ inwardly to it. Let $A_{1},\ B_{1},\ C_{1}$ be, the midpoints of the side-segments $BC,\ AC,\ AB$ respectively and also let $A_{2},\ B_{2},\ C_{2}$ be, the midpoints of the segments $AA',\ BB',\ CC'$ respectively. Prove that the line segments $A_{1}A_{2},\ B_{1}B_{2},\ C_{1}C_{2},$ are concurrent at one point.

PROOF. – It is clear that the points $A_{2},\ B_{2},\ C_{2},$ lie on $B_{1}C_{1},\ A_{1}C_{1},\ A_{1}B_{1}$ respectively and so, because of $B_{1}C_{1}\parallel BC,$

based on the Thales theorem, we have $\frac{A_{2}B_{1}}{A_{2}C_{1}} = \frac{A'C}{A'B}$ $,(1)$

Similarly we have $\frac{B_{2}C_{1}}{B_{2}A_{1}} = \frac{B'A}{B'C}$ $,(2)$ and $\frac{C_{2}A_{1}}{C_{2}B_{1}} = \frac{C'B}{C'A}$ $,(3)$

From $(1),$ $(2),$ $(3)$ $\Longrightarrow$ $\frac{A_{2}B_{1}}{A_{2}C_{1}}\cdot \frac{B_{2}C_{1}}{B_{2}A_{1}}\cdot \frac{C_{2}A_{1}}{C_{2}B_{1}} = \frac{A'C}{A'B}\cdot \f...$ $,(4)$

From $(4),$ based on the Ceva theorem, we conclude that the line segments $A_{1}A_{2},\ B_{1}B_{2},\ C_{1}C_{2},$ are concurrent at one point and the proof is completed.

Kostas Vittas.

**Posted:** Fri Aug 08, 2008 11:12 am

encyclopedia · Poincare Conjecture Offline Joined: 16 Apr 2008 Posts: 145

**Posted:** Sat Aug 09, 2008 7:35 am