- 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, CF be, it's altitudes.

We draw three circles (K1), (K2), (K3), with side segments BC, AC, AB

respectively, as diameters.

We denote as D', D'', the intersection points of line AD from

circle (K1). Similarly we denote the pair of points E', E'' and F',

F'', on lines BE, CF, respectively. ( The points D', E', F', inwardly

to the triangle ABC ).

We denote A', as the intersection point of the lines E'F'',

E''F'. Similarly B' as the one of the lines D'F'', D''F' and C' as

the one of the lines D'E'', D''E'.

Results :

1. The points A', B', C', lie on BC, AC, AB, respectively.

2. The lines AA', BB', CC', are concurrent at a point, so be it M.

3. The lines A'D',B'E',C'F', are concurrent at a point, so be it L.

4. The lines A'D'',B'E'',C'F'',are concurrent at a point,so be it N.

5. The points L, M, N, lie on a line pass through orthocenter H, of

the triangle ABC.

6. The points L, N, are the circumcenters of the triangles D'E'F',

D''E''F'', respectively.

7. If D1, E1, F1, are the intersection points of circles (K1), (K2),

(K3) respectively, from the circumcircle of the triangle

D''E''F'', then the lines AD1, BE1, CF1, are concurrent.

8. If D2, E2, C2, are the intersection points of circles (K1), (K2),

(K3) respectively, from the circumcircle of the triangle D'E'F',

then the lines D1D2, E1E2, F1F2, are concurrent at orthocenter H

of ABC.

I will try to post here next time, some proofs I have in mind for

(1), (2),...(6). I still not have proofs for (7), (8) and I am

waiting, with pleasure, for any idea. ( My drawing is erected on a

triangle ABC such that BC = 9.5, AC = 8.5, AB = 6.5 ).

Best regards.

Kostas Vittas. - 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) at X1, X2. Similarly we have (T2) and Y1, Y2; (T3) and Z1, Z2. Six points X1, X2, Y1, Y2, Z1, Z2 are concyclic on one circle centered at centroid G of triangle ABC.

Best regards,

Bui Quang Tuan

Kostas Vittas <vittasko@...> wrote:

...

A triangle ABC is given and let AD, BE, CF be, it's altitudes.

We draw three circles (K1), (K2), (K3), with side segments BC, AC, AB

respectively, as diameters.

We denote as D', D'', the intersection points of line AD from

circle (K1). Similarly we denote the pair of points E', E'' and F',

F'', on lines BE, CF, respectively. ( The points D', E', F', inwardly

to the triangle ABC ).

...

---------------------------------

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[Non-text portions of this message have been removed] - --- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>

wrote:

> Dear all my friends.

N.

>

> I would like to present an interesting configuration as

>follows : ( I apologize for my English )

>

> A triangle ABC is given and let AD, BE, CF be, it's altitudes.

> We draw three circles (K1), (K2), (K3), with side segments BC, AC,

>AB respectively, as diameters.

>

> We denote as D', D'', the intersection points of line AD from

> circle (K1). Similarly we denote the pair of points E', E'' and F',

> F'', on lines BE, CF, respectively. ( The points D', E', F',

>inwardly to the triangle ABC ).

>

> We denote A', as the intersection point of the lines E'F'',

> E''F'. Similarly B' as the one of the lines D'F'', D''F' and C' as

> the one of the lines D'E'', D''E'.

>

> Results :

>

> 1. The points A', B', C', lie on BC, AC, AB, respectively.

> 2. The lines AA', BB', CC', are concurrent at a point, so be it M.

> 3. The lines A'D',B'E',C'F', are concurrent at a point, so be it L.

> 4. The lines A'D'',B'E'',C'F'',are concurrent at a point,so be it

> 5. The points L, M, N, lie on a line pass through orthocenter H,

Dear all my friends.

> of the triangle ABC.

> 6. The points L, N, are the circumcenters of the triangles D'E'F',

> D''E''F'', respectively.

> 7. If D1, E1, F1, are the intersection points of circles (K1),

> (K2),(K3) respectively, from the circumcircle of the triangle

> D''E''F'', then the lines AD1, BE1, CF1, are concurrent.

> 8. If D2,E2,F2, are the intersection points of circles (K1),(K2),

> (K3)respectively, from the circumcircle of the triangle D'E'F',

> then the lines D1D2, E1E2, F1F2, are concurrent at orthocenter

> H of ABC.

>

> ...

I would like to present the proof of proposition (7), as

follows: ( my drawing is erected on a triangle with sides

BC = 7.5, AC = 6.8, AB = 5.2 )

7a. Let A' be the intersection point of E1E'',F1F'' and B', the

intersection point of D1D'', F1F''and C',the intersection point

of D1D'', E1E''.

7b. The point A' is the radical center of the circles (K2),(K3),(M)

( the circumcircle of the triangle D''E''F'' ). That is the

segment line AD, passes through the point A'. Similarly the

segment lines BE, CF, pass through the points B',C'respectively.

7c. Let R be the intersection point of BC, D1D''. We draw a line

perpendicular to the segment line BC at point R, which cuts the

segment lines BD1, CD1, at points Q, S respectively.

7d. The point C, is the orthocenter of the triangle BQS ( because of

the lines BR,SD1,are perpendicular to the lines QS,BQ

respectively ).Hence the segment lines QC, BS,are perpendicular

each other and their intersection point, so be it D'1, lies on

the circle (K1).

7e. We will prove that the point D'1 coincides to the point D'.

Really from the cyclic quadrilateral CD1QR we have that

<RCQ = <RD1Q (1). But <RCQ = <BCD'1 (2) and <RD1Q = <BD1D''(3).

From (1),(2),(3) => <BCD'1 = <BD1D'' => BD''= BD'1 (4)

Hence the point D'1 coincides to the point D'. That is, we have

that the segment line CD' passes through the point Q.

7f. As a helping proposition, we use the result that the points B',

A, S, are collinear ( we shall prove the proposition for this

result, below ). Hence, we consider the segments AD', BB' CC',

concurrent at orthocenter H of ABC and then, by the Lemma 2, we

have that the points Q' ( the intersection point of AC', CD' ),

R, S, as the intersection points of the opposite sides, the

other than H, of the quadrilaterals, AD'CC', BCC'B', AB'BD'

respectively, are collinear. Hence the point Q' coincides to

the point Q and then we have that the segment line AC', passes

through the point Q.

7g. Now, because of the points Q, R, S, are collinear, we conclude

that the triangles AB'C', BCD1, are perspective at one point, so

be it A1. That is the point A1, as the intersection point of

BC',B'C,lies on AD1. Similarly The point B1, as the intersection

point of AC',A'C and the point C1, as the intersection point of

AB', A'B, lie on BE1, CF1 respectively.

7h. We consider the segments AA',BB',CC', concurrent at orthocenter

H of ABC. So, by the Lemma 1, we have that the lines AA1, BB1,

CC1, are concurrent. Hence the lines AD1, BE1, CF1, are

concurrent at one point, so be it P and the proof of proposition

(7), is completed.

7.1 Helping Proposition. A triangle ABC is given. Let AD, BE,

CF be, it's altitudes and let H be, it's orthocenter. We draw

the circumcircle (K1) of the cyclic quadrilateral BDHF and we

denote as G, the intersection point of (K1), DE. Also we denote

as L, the intersection point of BE, FG. We consider an arbitrary

point between F, L and we denote as N, the intersection point of

(K1), HP.

If R, is the intersection point of BN, FG and Q, is the one

of BP, EG, prove that the points Q, R, A, are collinear.

7.1a Proof. We construct the line(e),through H and

perpendicular to the line BE. We denote as F', R', L', the

intersection points of HF, HR,(e)respectively, from the line BP.

7.1b It is easy to prove that the segment line FG,is

perpendicular to the BE ( from cyclic quadrilateral HDCE we

have that <EHC = <EDC, etc ). So, the point P, is the

orthocenter of the triangle RBH and then we have that the

segment line BR', is perpendicular to HR. Hence the point R',

lies on circle (K1) ( BH is diameter of (K1) ).

7.1c The points Q, R, A, lie on the side lines BP, FP, BF

respectively, of the triangle BFP. So, in order to these points

are collinear, it is enough to prove that is true, the

equality: [(RP):(RF)]·[(AF):(AB)]·[(QB):(QP)] = 1 (1)

( Menelaus theorem )

Because of the segment lines FL, AE are parallel,

=> (AF):(AB) = (EL): (EB) (2)

Also from the triangle BLP, by Menelaus theorem, we have that

=> [(EL):(EB)]·[(QB):(QP)]·[(GP):(GL)] = 1 (3)

By (2),(3),the equality (1) becomes:(RP): (RF)=(GP):(GL) (4)

From (4) => (RPRF):(RF)=(GPGL):(GL) => (FP):(RF)=(LP):(GL)

=> (FP):(RF) = (LP):(FL) (5) ( because of FL = GL )

Hence it is enough to prove that is true the equality (5),

instead of (1).

7.1d We compare the bundles of lines B.LPFR and H.L'R'F'P.

Their homologous lines, form congruent angles ( <LBP = <L'HR',

<PBF = <R'HF', <FBR = <F'HP ).

So, their cross-ratios are congruent. That is we have

that ( L,P,F,R ) = ( L',R',F',P ) (6)

Because of ( L',R',F',P ) = ( P,F',R',L' )

=> ( L,P,F,R ) = ( P,F',R',L' ) (7)

Also because of PR //(e) we have that

( P,F',R',L' )=( P,F,R ) (8)

From (7),(8),=> ( L,P,F,R ) = ( P,F,R ) (9)

From (9) => [(FL):(FP)]:[(RL):(RP)] = (RP):(RF)

=> [(FL):(FP)]·[(RP):(RL)] = (RP):(RF)

=> (FL):(FP) = (RL):(RF) => (FP):(RF) = (FL):(RL) (10)

7.1e From (5),(10),=>(LP):(FL)=(FL):(RL)=>(FL)²=(LP)·(RL) (11)

From the orthogonal triangle BFH ( <BFH = 90º )

=>(FL)²=(LB)·(LH) (12)

From (11),(12), => (LP)·(RL) = (LB)·(LH) (13)

Hence it is enough to prove that is true the equality (13),

instead of (1).

Really, this equality is true because of the similarity of the

orthogonal triangles, PLH and BLR ( <PLH = 90º, <BLR = 90º ).

We conclude now, that is also true the equality (1). Hence the

points Q, R, A, are collinear and the proof of proposition

(7.1), is completed.

Sometimes ( many times for me ), our mind follows a complicated

or a long way for the solution of a problem. So, I am waiting

for your advices, for a simpler synthetic proof.

I don't know if P ( the concurrency point of AD1, BE1, CF1 ), is

a well known point in the plane of a triangle and I still not

have studied, if it has any characteristic property.

I will post here, next time, the proofs of Lemmas 1, 2 ( see

mesage 13171 ) and also the ( easier ) proofs of propositions

(9), (10),...,(15).

I still not have any proof, for the proposition (8).

Best regards.

Kostas Vittas. - --- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>

wrote:> ...

Dear all my friends.

>

>7f. As a helping proposition, we use the result that the points B',

>A, S, are collinear ( we shall prove the proposition for this

>result, below ). Hence, we consider the segments ...

>

> ...

A friend of mine Kostas Helatiotis, said me that we don't need

any helping proposition for the proof of (7), in message 13274.

7f. From cyclic non-convex hexagon D'D''D1CEB, by Pascal's

theorem, we have that the points B', A, S, are collinear. Hence, we

consider the segments ...

I apologize for my blindness and please, cancel all about the

proof of the helping proposition (7.1), as unnecessary prattlings.

Best regards.

Kostas Vittas.