The results as you mentioned in your configuration, is true in

general case of a triangle A'B'C' and three arbitrary rectangles

B'C'CaBa, C'A'AbCb, A'B'BcAc, erected on sidesegments B'C', C'A',

A'B' respectively, outwardly (or inwardly) to it. So, we have again

the results as follows:

1). The lines A'A'', B'B'', C'C'', are concurrent at one point, so

be it P, as the homothetic center of A'B'C', A''B''C''.

2). The midperpendiculars of the segments AbAc, BaBc, CaCb, are

concurrent at one point, so be it Q.

2). The lines through A'', B'', C'', and perpendicular to AbAc,

BaBc, CaCb respectively, are concurrent at one point, so be it R.

3). The lines through A', B', C', and perpendicular to AbAc, BaBc,

CaCb respectively, are concurrent at one point, so be it R' (and so,

QR = QR').

4). The points P, Q, R, R', are collinear.

5). We draw two lines through Ba, Ca and perpendicular to BaBc,

CaCb respectively and we denote as A1, their intersection point.

Similarly we define the points B1, C1. So, we can prove that the

quadrilaterals B1C1AcAb, A1C1BcBa, A1B1CbCa, are rectangles and

hence, the point Q, is the circumcenter of the triangle A1B1C1.

But all the above results are well known. A very interesting work by

Nikos Dergiades and Flour van Lamoen, has already been posted in

journal Forumgeom. Please see at:

http://forumgeom.fau.edu/FG2003volume3/FG200316index.html

The proofs are given by barycentrics and it is a nice challenge, to

try someone for the solutions by classical geometry.

I have in mind some elementary proofs of the above results and with

pleasure, I will present here next time.

Best regards.

Kostas Vittas.

--- In Hyacinthos@yahoogroups.com, Quang Tuan Bui <bqtuan1962@...>

wrote:>

respectively. One line La passing through vertex A and parallel with

> Dear All My Friends,

> Given triangle ABC with points A', B', C' on sidelines BC, CA, AB

B'C'; Ba, Ca are orthogonal projections of B', C' on the line La

respectively. We have one rectangle B'C'CaBa. Similarly from vertices

B and C we can construct two other lines Lb, Lc and two other

rectangles: C'A'AbCb, A'B'BcAc. Three lines La, Lb, Lc bound one

triangle A''B''C''. It is clear that A''B''C'' and A'B'C' are

homothetic with homothetic center, say P.> Please prove that:

CaCb are concurrent at one point say Q.

> 1. Three perpendicular bisectors of three segments AbAc, BcBa,

> 2. Three perpendicular lines from A'', B'', C'' to AbAc, BcBa,

CaCb respectively are concurrent at one point say R.

> 3. Three points P, Q, R are collinear.

triangle of ABC?

> 4. What we can say about the ratio PQ/PR?

> 5. Are there somethings interesting when A'B'C' is Cevian

> Best regards,

and get things done faster.

> Bui Quang Tuan

>

>

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