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13274Re: Concurrencies on a trianle.

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  • Kostas Vittas
    Jun 13, 2006
      --- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>
      wrote:

      > 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,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.
      >
      > ...

      Dear all my friends.

      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) => (RP–RF):(RF)=(GP–GL):(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.
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