Dear all my friends. I would like to present an interesting configuration as follows : ( I apologize for my English ) A triangle ABC is given and let AD, BE,...
13162
Quang Tuan Bui
bqtuan1962@...
Jun 1, 2006 2:29 pm
Dear Kostas, You can see some results (only true if ABC is acute): E'E''F'F'' are concyclic on circle (T1) centered at A and orthogonal with (K1), cutting (K1)...
Dear Tuan and all. Thank you for your advices. It is true that the points E', E'', F', F'', are concyclic on circle centered at vertex A and we need this...
Dear all my friends. I post here the proofs of (1), (2) and I am waiting for your remarks. 1a. Because of AC is the diameter of circle (K2), we have AE'= AE''....
Dear Nikos, I seem to have come full circle in my thinking about this problem -- I agree that areal coordinates are the best to use. And, I really like your...
Dear Jeff, a direct proof is the following: If the triangles ABC, A#B#C# are perspective at P' = (p' : q' : r' ) in absolute barycentrics we would have: P = (p...
Dear all my friends. I post here the proofs of (3),(4),(5), beginning from (4). 4a. Because of the lines AA', BB', CC', are concurrent at point M, we have that...
13172
ddtm
ddtm@...
Jun 3, 2006 9:24 am
Dear friends! I offer you a theorem. Illustration: http://ddtm.liceum.ru/3-c-l.htm Let us consider an arbitrary quadrilateral ABCD. I' is the intersection of...
13173
ddtm
ddtm@...
Jun 3, 2006 9:25 am
Dear friends! I think, I've found some new points on Gauss line. Illustration: http://ddtm.liceum.ru/gauss.htm Points M[1], M[4] and M[8] were in original...
Dear all my friends. I post here the proof of (6) as follows : 6a. In the configuration of the circumscribed quadrilateral EaEbFaFb, with intouch quadrilateral...
13175
Quang Tuan Bui
bqtuan1962@...
Jun 3, 2006 4:38 pm
Dear All My Friends, Some weeks ago when working with my message "Euler and Feuerbach", I saw one interesting fact. In reference triangle there are three small...
Dear Yaroslav Ganin. Welcome to Hyakinthos. We can prove your proposition based on the follows simple idea : Lemma. If the orthogonal projections of three...
Dear Yaroslav. I would like to inform you for another very nice application of this Lemma, in the configuration of a bicentric quadrilateral. It is well known...
Dear Yaroslav. I mean when the corresponded points ( A, A' etc ), are coincided with the intersection point ( if intersect each other ) of the given lines...
Dear Yaroslav. I think that may be help us the idea which has been used for the solution of the `'Ganin theorem'' ( see messages 12466, 12499 ). In a quick...
Let ABC be a triangle, P = (x:y:z) a point and A'B'C' the pedal triangle of P. Denote Ab, Ac : the reflections of A in B',C' Bc, Ba : the reflections of B in...
Dear Yaroslav. It is clear that some of midpoints, as for example M[2],M[3], M[7], lie on Gauss line ( easy to prove by Thales theorem ). Best regards. Kostas...
Dear Yaroslav and Kostas, point M6 is the centroid of BEFD, hence the corresponding concurrence. All the other segments in question bisected by the Gauss line...
Dear Vladimir and Yaroslav. Thank you for your reminding that (so simple) the midpoint M[6] lies on the Gauss line, as the cedroid of the quadrilateral BDFE. ...
Dear Antreas, Nikolaos and Jeff, Here is a synthetic proof, based only on simple geometry, without a coordinate, matrix or theorem in sight, but, as alluded to...
13186
Quang Tuan Bui
bqtuan1962@...
Jun 4, 2006 11:00 pm
Dear All My Friends, I generalize my previous loci to one general case as following: Given triangle ABC; P is a Cevian point with Cevian triangle A'B'C'; X is...
Dear Peter, I do like using third dimension in plane geometry, but I can't understand your proof in the case P not = Q. ... Here I stop: if C1 lies on QC...
13188
Quang Tuan Bui
bqtuan1962@...
Jun 5, 2006 3:14 am
Dear Antreas and All, It is very interesting generalization! I see some thing may be help you to calculate the coordinates: 1. With any point P on the plane,...
13189
Quang Tuan Bui
bqtuan1962@...
Jun 5, 2006 4:13 am
Dear Antreas and All, Since three triangles AAbAc, BBcBa, CCaCb share one common circumcenter P (easy to show from 3) we can quick draw this configuration by...
[Antreas] ... [Bui Quang Tuan] ... Dear Tuan I think that, more generally, we have: Let ABC be a triangle and A'B'C' a triangle INSCRIBED in ABC. Denote Ab, Ac...
Dear Vladimir, Thanks for your comment and the query. There is nothing wrong. This technique of proof asks you to defer knowledge of certain properties of the...
[Antreas] ... We have B', C' are midpoints of AAb, AAc C', A' are midpoints of BBc, BBa A', B' are midpoints of CCa, CCb (The Nine Point Circle of AAbAc passes...
13193
Quang Tuan Bui
bqtuan1962@...
Jun 5, 2006 8:13 am
Dear Antreas, Yes, I see and try to study the general case. Now we have six cases: 1. General case: any INSCRIBED triangle A'B'C' (circle concurrency only) Ab,...
