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#1 h Feb 2, 2004, 1:06 PM • 2 Y

Y by Adventure10, Mango247

Let ABCD be the equilateral triangle and Ao,Bo,Co be the midpoints
of segments BC,CA,AB .The lines d1,d2,d3 each of which parallels two
others,through Ao,Bo,Co respectively , cut BoCo,CoAo,AoBo respectively
at A1,B1,C1 . Prove that AA1,BB1,CC1 are concurrent

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#2 h May 21, 2004, 7:17 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

There is a very powerful generalization (actually a classical theorem):

Cevian Nests Theorem. Let ABC be an arbitrary triangle, and A', B', C' three points on the sides BC, CA, AB such that the lines AA', BB', CC' concur. Let also A", B", C" be three points on the sides B'C', C'A', A'B' of triangle A'B'C' such that the lines A'A", B'B", C'C" concur. Then, the lines AA", BB", CC" concur, too.

In your case, A1lqdSchool, A', B', C' are the midpoints Ao, Bo, Co of BC, CA, AB (the lines AA', BB', CC' concur at the centroid of triangle ABC), and A", B", C" are the points A1, B1, C1 (the lines A'A", B'B", C'C" are parallel - i. e., we can say they concur at an infinite point). Hence your problem is a special case of the Cevian Nests Theorem.

Here is my favorite proof of the Cevian Nests Theorem:

We will use a lemma:

Menelaos Theorem for quadrilaterals. If X, Y, Z, W are points on the sidelines AB, BC, CD, DA of a quadrilateral ABCD, and these points X, Y, Z, W are collinear, then

AX / XB * BY / YC * CZ / ZD * DW / WA = 1

with directed segments.

This is very easy to prove. (Incidentally, the converse is not true, unlike the converse of the Menelaos theorem for triangles.)

Now, let the lines AA", BB", CC" meet the sides BC, CA, AB of the triangle ABC at the points X, Y, Z, respectively. In order to prove the Cevian Nests Theorem, we have to show that the lines AA", BB", CC" concur. This will be clear by Ceva once we succeed to show

BX / XC * CY / YA * AZ / ZB = 1.

Well, apply the Menelaos Theorem for quadrilaterals to the quadrilateral BCB'C' with the collinear points X, A, A", A on its sides (you are reading right - the point A occurs twice, it is actually the point of intersection of two sides of the quadrilateral!). Then, you obtain

BX / XC * CA / AB' * B'A" / A"C' * C'A / AB = 1,

so that

BX / XC = AB / C'A * A"C' / B'A" * AB' / CA
= (AB' / C'A) * (AB / CA) * (A"C' / B'A").

Similarly,

CY / YA = (BC' / A'B) * (BC / AB) * (B"A' / C'B"), and
AZ / ZB = (CA' / B'C) * (CA / BC) * (C"B' / A'C").

Hence,

BX / XC * CY / YA * AZ / ZB
= (AB' / C'A) * (AB / CA) * (A"C' / B'A")
* (BC' / A'B) * (BC / AB) * (B"A' / C'B")
* (CA' / B'C) * (CA / BC) * (C"B' / A'C")
= (AB' / C'A * BC' / A'B * CA' / B'C)
* (AB / CA * BC / AB * CA / BC)
* (A"C' / B'A" * B"A' / C'B" * C"B' / A'C").

Now, all three bracketed products are = 1, because after Ceva, the concurrent lines AA', BB', CC' yield

AB' / C'A * BC' / A'B * CA' / B'C = 1,

and the concurrent lines A'A", B'B", C'C" yield

A"C' / B'A" * B"A' / C'B" * C"B' / A'C" = 1,

and finally, AB / CA * BC / AB * CA / BC = 1 is obvious. Hence, you have

BX / XC * CY / YA * AZ / ZB = 1,

and after Ceva, the lines AX, BY, CZ concur, i. e. the lines AA", BB", CC" concur. Qed..

See http://www.mathlinks.ro/Forum/viewtopic.php?t=6579 for an extension of the Cevian Nests Theorem with a proof (which is actually the above proof after a few modifications).

Darij

This post has been edited 2 times. Last edited by darij grinberg, Dec 24, 2007, 8:57 PM

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#3 h Jun 1, 2004, 8:45 PM • 2 Y

Y by Adventure10, Mango247

This would be my favorite proof for the cevian nests theorem.
It's a projective geometry proof.
Take the line joining the intersection points of the 2 sets of concurent cevinas to infinity to get the picture bellow.

In the picture, AA',BB',CC' are parallel. So are A'A",B'B" and C'C".
If you can prove that AA", BB" and CC" are concurent, you solve the problem.

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