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Re: ÈÝìá: [EMHL] Re: Circles in cevians

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  • Quang Tuan Bui
    Dear Fred, You are right. I made one typo mistake when wrote INCIRCLE. It should be NINE POINT CIRCLE. Sorry for mistake! Best regards, Bui Quang Tuan
    Message 1 of 26 , Mar 31, 2007
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      Dear Fred,
      You are right. I made one typo mistake when wrote INCIRCLE.
      It should be NINE POINT CIRCLE.
      Sorry for mistake!
      Best regards,
      Bui Quang Tuan


      fredlangch <fred.lang@...> wrote: Dear all,
      I obtain the same results with the following complements:

      1) the quintic of the first locus
      passes by A,B,C, the vertices of the anticomplementary triangle and te infinite points of
      the sides of ABC

      2) the second locus is of degree 9:
      infinite line (twice), circumcircle, nine points circle, the cubic K043

      3) a third locus is the set of P such taht H1H2H3 and L1L2L3 are concurrents.
      I found a degree 18 locus:
      the three median lines of ABC, infinite line (four time),nine points circle, a quartic and
      another quintic.
      On the quartic, I found the centers X(110) and X(265) and the points ABC
      On the quintic I found the same points as the quintic of the first locus.

      4) locus of perspectors???

      Friendly
      Fred

      --- In Hyacinthos@yahoogroups.com, Quang Tuan Bui wrote:
      >
      > Dear Nikos and Jean-Pierre,
      > There is another may be interesting locus:
      > If Ma, Mb, Mc are midpoints of BC, CA, AB resp. and H1, H2, H3 are circumcenters of
      PMbMc, PMcMa, PMaMb resp. then locus of P such that two triangles K1K2K3 and H1H2H3
      are perspective is union of:
      > - Infinite line
      > - Circumcircle
      > - Incircle
      > - K043 cubic
      > Best regards,
      > Bui Quang Tuan
      >
      >
      > Nikolaos Dergiades wrote: Dear Jean-Pierre and Tuan,
      >
      > [JPE]
      > > If we use any inversion with center P, this is
      > > Desargues theorem.
      >
      >
      >
      > Very good, proof.
      >
      > > It seems to be the circumcircle of the
      > > anticomplementary triangle
      >
      > [QTB]
      > > - K1L1, K2L2, K3L3 are concurrent (I think we can
      > >find synthetic
      > >proof for this fact)
      > > - A1, B1, C1 = reflections of P in K1L1, K2L2,
      > >K3L3 so P, A1, B1, C1
      > >are concyclic.
      > > - K1K2 = L1L2 if and only if K2K3 = L2L3 (if and
      > >only if K3K1 = L3L1)
      > > - Part locus conic is circumcircle of anti
      > >complementary triangle.
      > > - Other part locus is one quintic
      >
      > Yes the conic is the circumcircle of
      > anticomplementary triangle.
      > The locus is also a quintic.
      > The triangles K1K2K3, L1L2L3 are homothetic
      > and the center of homothety is the center
      > of the circle of PA1B1C1.
      > Since the triangles are homothetic they
      > are congruent if they have equal areas.
      >
      > Best regards
      > Nikos Dergiades
      >
      >


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    • Vladimir Dubrovsky
      Dear Tuan, Ricardo, Kostas, Tarik, Francois and Nikolaos Somehow I missed some posts and found another solution to Tuan s question. It seems to be shorter, so
      Message 2 of 26 , Apr 2, 2007
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        Dear Tuan, Ricardo, Kostas, Tarik, Francois and Nikolaos

        Somehow I missed some posts and found another solution to Tuan's question.
        It seems to be shorter, so I decided to add it to the collection.

        So we have point P on AD such that Angle BPD=Angle BAC=2Angle DPC.

        Extend AD to meet the circumcircle of ABC at E; let CE=d, BE=e.
        Then
        BD:DC=area(ABE):area(ACE)=ce/bd. (*)

        Triangle PBE is similar to ABC; hence PE=be/a. (This is, actually,
        Francois's similarity.)

        Triangle CPF is similar to A'AB, where AA' is the bisector of A (this is
        another similarity, but a transformational argument escapes me; maybe
        Francois can shed light on it).
        Hence PE= cd/BA' =(b+c)d/a.
        It remains to equate the two expressions for PE and substitute the resulting
        e/d=(b+c)/b into (*).

        Notice that if we take triangle EBC as the initial one (instead of ABC),
        then P will be on the *extension* of ED. The original relation between the
        angles is violated, but it will remain the same if we think of angles as
        oriented angles between lines rather than rays. So, in a certain way, this
        answers Tarik's question: BD/DC = e(e-d)/dd.

        Best regards,
        Vladimir

        ==============

        [QTB]
        >> This fact is still true for any triangle ABC. In
        >> this case, D divide BC by one simple ratio depending
        >> on sides a, b, c. What is this ratio and how is the
        >> proof for this general case?
        >
        [TA]
        > Dear kostas,
        > if the point P is on the extension of the line AD
        > will your proof work there?
        >Moon Bangladesh
        >
        [ND]
        > We construct a point D on BC such that
        > BD/DC = k =c(b+c)/bb
        > The parallel from D to AC meets AB at E.
        > The circumcircle EBC meets AC at F.
        > The circumcircle EBD meets AD at P.
        > It is easy to prove that DF is parallel to
        > AA' the A_bisector
        > from CA'=ac/(b+c) DC=a/(k+1) AE = c/(k+1)
        > AE.AB = AF.AC and CF/CA = CD/CA'
        > Hence <BPD = <BED = <BAC = 2 <DPC = 2 <DFC
        >
        > Hence the ratio is k. For b=c we get k=2
        > as in Barosso's problem.
        >
        > The contruction of D is as follows:
        > We costruct the parallelogram BCAC1.
        > The parallel from C to AA' meets AB at C2
        > The reflection of A in B is the point C3.
        > The circumcircle of C1C2C3 meets the line
        > BC1 at C4. The line AC4 meets BC at the
        > point we want D.
        >
        > Best regards
        > Nikos Dergiades

        [FR]
        > Here another proof where I have tried to minimize the number of auxiliary
        > points.
        >
        > Let f be the direct similarity of center B sending A to P and E = f(C).
        > By hypothesis, A, P, E are on a same line.
        > We have the following equalities between angles of lines:
        > (CA, CB) = (EP, EB) (invariance of angles by a direct similarity)
        > (EP, EB) = (EA, EB) (for A, B, P are on a same line)
        > So (EA, EB) = (CA, CB) and points A, B, C, E are cocyclic (i.e on a same
        > circle)
        >
        > Now as A is on the perpendicular bisector of segment BC, line EA is a
        > bisector of angle <BEC.
        > The other bisector is through the point D" harmonic conjugate of D wrt BC.
        > But by assumption done on D (i.e BD = 2 DC), C is the middle of BD".
        >
        > Let F be the middle of segment BE. Then FC is parallel to ED"and hence
        > perpendicular to line EDPA.
        > Hence C and F are symmetric wrt this last line and we are done :
        > <FPE = <EPC = (1/2) <BAC
        >
        > Un abrazo
        > Francois
      • Quang Tuan Bui
        Dear All My Friends, Thank you very much for your interesting proofs and remarks. Thanks to Ricardo Barroso! Please inform us from where you have got this
        Message 3 of 26 , Apr 3, 2007
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          Dear All My Friends,

          Thank you very much for your interesting proofs and remarks.
          Thanks to Ricardo Barroso! Please inform us from where you have got this problem!

          I would like to post here my original construction when I found this general fact for complete collection. I use here my notations.
          - Internal angle bisector of A cuts BC at A'
          - Parallel line through A' with AC cuts AB at C'
          - Parallel line through A' with AB cuts AC at B'
          - A* = intersection of BB', CC'
          - Parallel line through A* with AC cuts BC at D, cuts AB at E
          - Circumcircle BDE cuts AD at P (other than D)
          - Parallel line through A* with AB cuts BC at D', cuts AC at E'
          - Circumcircle CD'E' cuts AD' at P' (other than D')
          We have similar results with two point P, P':
          Angle(BPD) = Angle(CP'D') = Angle(BAC) = 2*Angle(DPC) = 2*Angle(D'P'B)
          We can see following concyclic point sets:
          - D, D', P, P'
          - D, D', D, E'
          - B, C, E, E'
          - C, D, P, E'
          - B, D', P', E
          - A, E, E', P, P'
          We can see also following two concurrencies:
          - AD, CC', A'B' are concurrent at X
          - AD', BB', A'C' are concurrent at Y
          and four points P, P', X, Y are also concyclic.
          If we construct all D, D' points for BC, CA, AB sides then we have total six D-points. These six D-points are on one conic. The D-hexagon has one property: three lines connected midpoints of opposite sides (pairwise) are concurrent. The concurrent point has not complicated barycentrics but not nice.
          Thank you again and best regards,
          Bui Quang Tuan


          Vladimir Dubrovsky <v_dubrovsky@...> wrote: Dear Tuan, Ricardo, Kostas, Tarik, Francois and Nikolaos

          Somehow I missed some posts and found another solution to Tuan's question.
          It seems to be shorter, so I decided to add it to the collection.

          So we have point P on AD such that Angle BPD=Angle BAC=2Angle DPC.

          Extend AD to meet the circumcircle of ABC at E; let CE=d, BE=e.
          Then
          BD:DC=area(ABE):area(ACE)=ce/bd. (*)

          Triangle PBE is similar to ABC; hence PE=be/a. (This is, actually,
          Francois's similarity.)

          Triangle CPF is similar to A'AB, where AA' is the bisector of A (this is
          another similarity, but a transformational argument escapes me; maybe
          Francois can shed light on it).
          Hence PE= cd/BA' =(b+c)d/a.
          It remains to equate the two expressions for PE and substitute the resulting
          e/d=(b+c)/b into (*).

          Notice that if we take triangle EBC as the initial one (instead of ABC),
          then P will be on the *extension* of ED. The original relation between the
          angles is violated, but it will remain the same if we think of angles as
          oriented angles between lines rather than rays. So, in a certain way, this
          answers Tarik's question: BD/DC = e(e-d)/dd.

          Best regards,
          Vladimir

          ==============

          [QTB]
          >> This fact is still true for any triangle ABC. In
          >> this case, D divide BC by one simple ratio depending
          >> on sides a, b, c. What is this ratio and how is the
          >> proof for this general case?
          >
          [TA]
          > Dear kostas,
          > if the point P is on the extension of the line AD
          > will your proof work there?
          >Moon Bangladesh
          >
          [ND]
          > We construct a point D on BC such that
          > BD/DC = k =c(b+c)/bb
          > The parallel from D to AC meets AB at E.
          > The circumcircle EBC meets AC at F.
          > The circumcircle EBD meets AD at P.
          > It is easy to prove that DF is parallel to
          > AA' the A_bisector
          > from CA'=ac/(b+c) DC=a/(k+1) AE = c/(k+1)
          > AE.AB = AF.AC and CF/CA = CD/CA'
          > Hence >
          > Hence the ratio is k. For b=c we get k=2
          > as in Barosso's problem.
          >
          > The contruction of D is as follows:
          > We costruct the parallelogram BCAC1.
          > The parallel from C to AA' meets AB at C2
          > The reflection of A in B is the point C3.
          > The circumcircle of C1C2C3 meets the line
          > BC1 at C4. The line AC4 meets BC at the
          > point we want D.
          >
          > Best regards
          > Nikos Dergiades

          [FR]
          > Here another proof where I have tried to minimize the number of auxiliary
          > points.
          >
          > Let f be the direct similarity of center B sending A to P and E = f(C).
          > By hypothesis, A, P, E are on a same line.
          > We have the following equalities between angles of lines:
          > (CA, CB) = (EP, EB) (invariance of angles by a direct similarity)
          > (EP, EB) = (EA, EB) (for A, B, P are on a same line)
          > So (EA, EB) = (CA, CB) and points A, B, C, E are cocyclic (i.e on a same
          > circle)
          >
          > Now as A is on the perpendicular bisector of segment BC, line EA is a
          > bisector of angle > The other bisector is through the point D" harmonic conjugate of D wrt BC.
          > But by assumption done on D (i.e BD = 2 DC), C is the middle of BD".
          >
          > Let F be the middle of segment BE. Then FC is parallel to ED"and hence
          > perpendicular to line EDPA.
          > Hence C and F are symmetric wrt this last line and we are done :
          > >
          > Un abrazo
          > Francois




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        • Quang Tuan Bui
          Dear All My Friends, Sorry for one typo mistake: The second concyclic set should be read truly as: D, D , E, E Best regards, Bui Quang Tuan Quang Tuan Bui
          Message 4 of 26 , Apr 3, 2007
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            Dear All My Friends,
            Sorry for one typo mistake:
            The second concyclic set should be read truly as: D, D', E, E'
            Best regards,
            Bui Quang Tuan

            Quang Tuan Bui <bqtuan1962@...> wrote:
            Dear All My Friends,

            Thank you very much for your interesting proofs and remarks.
            Thanks to Ricardo Barroso! Please inform us from where you have got this problem!

            I would like to post here my original construction when I found this general fact for complete collection. I use here my notations.
            - Internal angle bisector of A cuts BC at A'
            - Parallel line through A' with AC cuts AB at C'
            - Parallel line through A' with AB cuts AC at B'
            - A* = intersection of BB', CC'
            - Parallel line through A* with AC cuts BC at D, cuts AB at E
            - Circumcircle BDE cuts AD at P (other than D)
            - Parallel line through A* with AB cuts BC at D', cuts AC at E'
            - Circumcircle CD'E' cuts AD' at P' (other than D')
            We have similar results with two point P, P':
            Angle(BPD) = Angle(CP'D') = Angle(BAC) = 2*Angle(DPC) = 2*Angle(D'P'B)
            We can see following concyclic point sets:
            - D, D', P, P'
            - D, D', D, E'
            - B, C, E, E'
            - C, D, P, E'
            - B, D', P', E
            - A, E, E', P, P'
            We can see also following two concurrencies:
            - AD, CC', A'B' are concurrent at X
            - AD', BB', A'C' are concurrent at Y
            and four points P, P', X, Y are also concyclic.
            If we construct all D, D' points for BC, CA, AB sides then we have total six D-points. These six D-points are on one conic. The D-hexagon has one property: three lines connected midpoints of opposite sides (pairwise) are concurrent. The concurrent point has not complicated barycentrics but not nice.
            Thank you again and best regards,
            Bui Quang Tuan


            Vladimir Dubrovsky wrote: Dear Tuan, Ricardo, Kostas, Tarik, Francois and Nikolaos

            Somehow I missed some posts and found another solution to Tuan's question.
            It seems to be shorter, so I decided to add it to the collection.

            So we have point P on AD such that Angle BPD=Angle BAC=2Angle DPC.

            Extend AD to meet the circumcircle of ABC at E; let CE=d, BE=e.
            Then
            BD:DC=area(ABE):area(ACE)=ce/bd. (*)

            Triangle PBE is similar to ABC; hence PE=be/a. (This is, actually,
            Francois's similarity.)

            Triangle CPF is similar to A'AB, where AA' is the bisector of A (this is
            another similarity, but a transformational argument escapes me; maybe
            Francois can shed light on it).
            Hence PE= cd/BA' =(b+c)d/a.
            It remains to equate the two expressions for PE and substitute the resulting
            e/d=(b+c)/b into (*).

            Notice that if we take triangle EBC as the initial one (instead of ABC),
            then P will be on the *extension* of ED. The original relation between the
            angles is violated, but it will remain the same if we think of angles as
            oriented angles between lines rather than rays. So, in a certain way, this
            answers Tarik's question: BD/DC = e(e-d)/dd.

            Best regards,
            Vladimir

            ==============

            [QTB]
            >> This fact is still true for any triangle ABC. In
            >> this case, D divide BC by one simple ratio depending
            >> on sides a, b, c. What is this ratio and how is the
            >> proof for this general case?
            >
            [TA]
            > Dear kostas,
            > if the point P is on the extension of the line AD
            > will your proof work there?
            >Moon Bangladesh
            >
            [ND]
            > We construct a point D on BC such that
            > BD/DC = k =c(b+c)/bb
            > The parallel from D to AC meets AB at E.
            > The circumcircle EBC meets AC at F.
            > The circumcircle EBD meets AD at P.
            > It is easy to prove that DF is parallel to
            > AA' the A_bisector
            > from CA'=ac/(b+c) DC=a/(k+1) AE = c/(k+1)
            > AE.AB = AF.AC and CF/CA = CD/CA'
            > Hence >
            > Hence the ratio is k. For b=c we get k=2
            > as in Barosso's problem.
            >
            > The contruction of D is as follows:
            > We costruct the parallelogram BCAC1.
            > The parallel from C to AA' meets AB at C2
            > The reflection of A in B is the point C3.
            > The circumcircle of C1C2C3 meets the line
            > BC1 at C4. The line AC4 meets BC at the
            > point we want D.
            >
            > Best regards
            > Nikos Dergiades

            [FR]
            > Here another proof where I have tried to minimize the number of auxiliary
            > points.
            >
            > Let f be the direct similarity of center B sending A to P and E = f(C).
            > By hypothesis, A, P, E are on a same line.
            > We have the following equalities between angles of lines:
            > (CA, CB) = (EP, EB) (invariance of angles by a direct similarity)
            > (EP, EB) = (EA, EB) (for A, B, P are on a same line)
            > So (EA, EB) = (CA, CB) and points A, B, C, E are cocyclic (i.e on a same
            > circle)
            >
            > Now as A is on the perpendicular bisector of segment BC, line EA is a
            > bisector of angle > The other bisector is through the point D" harmonic conjugate of D wrt BC.
            > But by assumption done on D (i.e BD = 2 DC), C is the middle of BD".
            >
            > Let F be the middle of segment BE. Then FC is parallel to ED"and hence
            > perpendicular to line EDPA.
            > Hence C and F are symmetric wrt this last line and we are done :
            > >
            > Un abrazo
            > Francois



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            Have a HUGE year through Yahoo! Small Business.

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          • Francois Rideau
            Dear friends Here another construction of point P, (another, I hope for I don t read very carefully other posts!) 1° I construct the circle Gamma on A and B,
            Message 5 of 26 , Apr 3, 2007
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              Dear friends
              Here another construction of point P, (another, I hope for I don't read very
              carefully other posts!)

              1° I construct the circle Gamma on A and B, tangent at A to line AC.
              2° The perpendicular bissector of segment BC cuts Gamma in U and U' such
              that AU is the internal bissector of <BAC and AU' is the other (external)
              bissector of the same angle <BAC.
              3° The parallel to line AC through U cuts again Gamma in F.
              4° Line FC cuts again circle Gamma in the sought point P
              Friendly
              Francois
              PD
              It remains to see what point P' is obtained when we perform the same
              construction with U', that is to say P' is the other intersection of line
              F'C with Gamma where U'F' is still parallel to line AC.
              Maybe we have 2 solutions when one uses directed angles of lines rather than
              "cheese angles"?


              On 03 Apr 2007 00:49:35 -0700, Quang Tuan Bui <bqtuan1962@...> wrote:
              >
              > Dear All My Friends,
              > Sorry for one typo mistake:
              > The second concyclic set should be read truly as: D, D', E, E'
              > Best regards,
              > Bui Quang Tuan
              >
              > Quang Tuan Bui <bqtuan1962@... <bqtuan1962%40yahoo.com>> wrote:
              > Dear All My Friends,
              >
              > Thank you very much for your interesting proofs and remarks.
              > Thanks to Ricardo Barroso! Please inform us from where you have got this
              > problem!
              >
              > I would like to post here my original construction when I found this
              > general fact for complete collection. I use here my notations.
              > - Internal angle bisector of A cuts BC at A'
              > - Parallel line through A' with AC cuts AB at C'
              > - Parallel line through A' with AB cuts AC at B'
              > - A* = intersection of BB', CC'
              > - Parallel line through A* with AC cuts BC at D, cuts AB at E
              > - Circumcircle BDE cuts AD at P (other than D)
              > - Parallel line through A* with AB cuts BC at D', cuts AC at E'
              > - Circumcircle CD'E' cuts AD' at P' (other than D')
              > We have similar results with two point P, P':
              > Angle(BPD) = Angle(CP'D') = Angle(BAC) = 2*Angle(DPC) = 2*Angle(D'P'B)
              > We can see following concyclic point sets:
              > - D, D', P, P'
              > - D, D', D, E'
              > - B, C, E, E'
              > - C, D, P, E'
              > - B, D', P', E
              > - A, E, E', P, P'
              > We can see also following two concurrencies:
              > - AD, CC', A'B' are concurrent at X
              > - AD', BB', A'C' are concurrent at Y
              > and four points P, P', X, Y are also concyclic.
              > If we construct all D, D' points for BC, CA, AB sides then we have total
              > six D-points. These six D-points are on one conic. The D-hexagon has one
              > property: three lines connected midpoints of opposite sides (pairwise) are
              > concurrent. The concurrent point has not complicated barycentrics but not
              > nice.
              > Thank you again and best regards,
              > Bui Quang Tuan
              >
              > Vladimir Dubrovsky wrote: Dear Tuan, Ricardo, Kostas, Tarik, Francois and
              > Nikolaos
              >
              > Somehow I missed some posts and found another solution to Tuan's question.
              >
              > It seems to be shorter, so I decided to add it to the collection.
              >
              > So we have point P on AD such that Angle BPD=Angle BAC=2Angle DPC.
              >
              > Extend AD to meet the circumcircle of ABC at E; let CE=d, BE=e.
              > Then
              > BD:DC=area(ABE):area(ACE)=ce/bd. (*)
              >
              > Triangle PBE is similar to ABC; hence PE=be/a. (This is, actually,
              > Francois's similarity.)
              >
              > Triangle CPF is similar to A'AB, where AA' is the bisector of A (this is
              > another similarity, but a transformational argument escapes me; maybe
              > Francois can shed light on it).
              > Hence PE= cd/BA' =(b+c)d/a.
              > It remains to equate the two expressions for PE and substitute the
              > resulting
              > e/d=(b+c)/b into (*).
              >
              > Notice that if we take triangle EBC as the initial one (instead of ABC),
              > then P will be on the *extension* of ED. The original relation between the
              >
              > angles is violated, but it will remain the same if we think of angles as
              > oriented angles between lines rather than rays. So, in a certain way, this
              >
              > answers Tarik's question: BD/DC = e(e-d)/dd.
              >
              > Best regards,
              > Vladimir
              >
              > ==============
              >
              > [QTB]
              > >> This fact is still true for any triangle ABC. In
              > >> this case, D divide BC by one simple ratio depending
              > >> on sides a, b, c. What is this ratio and how is the
              > >> proof for this general case?
              > >
              > [TA]
              > > Dear kostas,
              > > if the point P is on the extension of the line AD
              > > will your proof work there?
              > >Moon Bangladesh
              > >
              > [ND]
              > > We construct a point D on BC such that
              > > BD/DC = k =c(b+c)/bb
              > > The parallel from D to AC meets AB at E.
              > > The circumcircle EBC meets AC at F.
              > > The circumcircle EBD meets AD at P.
              > > It is easy to prove that DF is parallel to
              > > AA' the A_bisector
              > > from CA'=ac/(b+c) DC=a/(k+1) AE = c/(k+1)
              > > AE.AB = AF.AC and CF/CA = CD/CA'
              > > Hence >
              > > Hence the ratio is k. For b=c we get k=2
              > > as in Barosso's problem.
              > >
              > > The contruction of D is as follows:
              > > We costruct the parallelogram BCAC1.
              > > The parallel from C to AA' meets AB at C2
              > > The reflection of A in B is the point C3.
              > > The circumcircle of C1C2C3 meets the line
              > > BC1 at C4. The line AC4 meets BC at the
              > > point we want D.
              > >
              > > Best regards
              > > Nikos Dergiades
              >
              > [FR]
              > > Here another proof where I have tried to minimize the number of
              > auxiliary
              > > points.
              > >
              > > Let f be the direct similarity of center B sending A to P and E = f(C).
              > > By hypothesis, A, P, E are on a same line.
              > > We have the following equalities between angles of lines:
              > > (CA, CB) = (EP, EB) (invariance of angles by a direct similarity)
              > > (EP, EB) = (EA, EB) (for A, B, P are on a same line)
              > > So (EA, EB) = (CA, CB) and points A, B, C, E are cocyclic (i.e on a same
              > > circle)
              > >
              > > Now as A is on the perpendicular bisector of segment BC, line EA is a
              > > bisector of angle > The other bisector is through the point D" harmonic
              > conjugate of D wrt BC.
              > > But by assumption done on D (i.e BD = 2 DC), C is the middle of BD".
              > >
              > > Let F be the middle of segment BE. Then FC is parallel to ED"and hence
              > > perpendicular to line EDPA.
              > > Hence C and F are symmetric wrt this last line and we are done :
              > > >
              > > Un abrazo
              > > Francois
              >
              > ---------------------------------
              > Get your own web address.
              > Have a HUGE year through Yahoo! Small Business.
              >
              > [Non-text portions of this message have been removed]
              >
              >
              >


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            • Ricardo Barroso
              Dear all my friends: The problem angles is from Mathematical Excalibur Vol 7 nº 3, problem 158 http://www.math.ust.hk/excalibur/v7_n3.pdf Best regard Ricardo
              Message 6 of 26 , Apr 3, 2007
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                Dear all my friends:

                The problem angles is from

                Mathematical Excalibur
                Vol 7 nº 3, problem 158

                http://www.math.ust.hk/excalibur/v7_n3.pdf


                Best regard



                Ricardo


                ----- Mensaje original ----
                De: Quang Tuan Bui <bqtuan1962@...>
                Para: Hyacinthos@yahoogroups.com
                Enviado: martes, 3 de abril, 2007 9:49:35
                Asunto: Re: [EMHL] Re: angles













                Dear All My Friends,

                Sorry for one typo mistake:

                The second concyclic set should be read truly as: D, D', E, E'

                Best regards,

                Bui Quang Tuan



                Quang Tuan Bui <bqtuan1962@yahoo. com> wrote:

                Dear All My Friends,



                Thank you very much for your interesting proofs and remarks.

                Thanks to Ricardo Barroso! Please inform us from where you have got this problem!



                I would like to post here my original construction when I found this general fact for complete collection. I use here my notations.

                - Internal angle bisector of A cuts BC at A'

                - Parallel line through A' with AC cuts AB at C'

                - Parallel line through A' with AB cuts AC at B'

                - A* = intersection of BB', CC'

                - Parallel line through A* with AC cuts BC at D, cuts AB at E

                - Circumcircle BDE cuts AD at P (other than D)

                - Parallel line through A* with AB cuts BC at D', cuts AC at E'

                - Circumcircle CD'E' cuts AD' at P' (other than D')

                We have similar results with two point P, P':

                Angle(BPD) = Angle(CP'D') = Angle(BAC) = 2*Angle(DPC) = 2*Angle(D'P' B)

                We can see following concyclic point sets:

                - D, D', P, P'

                - D, D', D, E'

                - B, C, E, E'

                - C, D, P, E'

                - B, D', P', E

                - A, E, E', P, P'

                We can see also following two concurrencies:

                - AD, CC', A'B' are concurrent at X

                - AD', BB', A'C' are concurrent at Y

                and four points P, P', X, Y are also concyclic.

                If we construct all D, D' points for BC, CA, AB sides then we have total six D-points. These six D-points are on one conic. The D-hexagon has one property: three lines connected midpoints of opposite sides (pairwise) are concurrent. The concurrent point has not complicated barycentrics but not nice.

                Thank you again and best regards,

                Bui Quang Tuan



                Vladimir Dubrovsky wrote: Dear Tuan, Ricardo, Kostas, Tarik, Francois and Nikolaos



                Somehow I missed some posts and found another solution to Tuan's question.

                It seems to be shorter, so I decided to add it to the collection.



                So we have point P on AD such that Angle BPD=Angle BAC=2Angle DPC.



                Extend AD to meet the circumcircle of ABC at E; let CE=d, BE=e.

                Then

                BD:DC=area(ABE) :area(ACE) =ce/bd. (*)



                Triangle PBE is similar to ABC; hence PE=be/a. (This is, actually,

                Francois's similarity.)



                Triangle CPF is similar to A'AB, where AA' is the bisector of A (this is

                another similarity, but a transformational argument escapes me; maybe

                Francois can shed light on it).

                Hence PE= cd/BA' =(b+c)d/a.

                It remains to equate the two expressions for PE and substitute the resulting

                e/d=(b+c)/b into (*).



                Notice that if we take triangle EBC as the initial one (instead of ABC),

                then P will be on the *extension* of ED. The original relation between the

                angles is violated, but it will remain the same if we think of angles as

                oriented angles between lines rather than rays. So, in a certain way, this

                answers Tarik's question: BD/DC = e(e-d)/dd.



                Best regards,

                Vladimir



                ============ ==



                [QTB]

                >> This fact is still true for any triangle ABC. In

                >> this case, D divide BC by one simple ratio depending

                >> on sides a, b, c. What is this ratio and how is the

                >> proof for this general case?

                >

                [TA]

                > Dear kostas,

                > if the point P is on the extension of the line AD

                > will your proof work there?

                >Moon Bangladesh

                >

                [ND]

                > We construct a point D on BC such that

                > BD/DC = k =c(b+c)/bb

                > The parallel from D to AC meets AB at E.

                > The circumcircle EBC meets AC at F.

                > The circumcircle EBD meets AD at P.

                > It is easy to prove that DF is parallel to

                > AA' the A_bisector

                > from CA'=ac/(b+c) DC=a/(k+1) AE = c/(k+1)

                > AE.AB = AF.AC and CF/CA = CD/CA'

                > Hence >

                > Hence the ratio is k. For b=c we get k=2

                > as in Barosso's problem.

                >

                > The contruction of D is as follows:

                > We costruct the parallelogram BCAC1.

                > The parallel from C to AA' meets AB at C2

                > The reflection of A in B is the point C3.

                > The circumcircle of C1C2C3 meets the line

                > BC1 at C4. The line AC4 meets BC at the

                > point we want D.

                >

                > Best regards

                > Nikos Dergiades



                [FR]

                > Here another proof where I have tried to minimize the number of auxiliary

                > points.

                >

                > Let f be the direct similarity of center B sending A to P and E = f(C).

                > By hypothesis, A, P, E are on a same line.

                > We have the following equalities between angles of lines:

                > (CA, CB) = (EP, EB) (invariance of angles by a direct similarity)

                > (EP, EB) = (EA, EB) (for A, B, P are on a same line)

                > So (EA, EB) = (CA, CB) and points A, B, C, E are cocyclic (i.e on a same

                > circle)

                >

                > Now as A is on the perpendicular bisector of segment BC, line EA is a

                > bisector of angle > The other bisector is through the point D" harmonic conjugate of D wrt BC.

                > But by assumption done on D (i.e BD = 2 DC), C is the middle of BD".

                >

                > Let F be the middle of segment BE. Then FC is parallel to ED"and hence

                > perpendicular to line EDPA.

                > Hence C and F are symmetric wrt this last line and we are done :

                > >

                > Un abrazo

                > Francois



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              • Ignacio Larrosa Cañestro
                ... You meant the perpendicular bissector of segment AB, of course ... Best regards, Ignacio Larrosa Cañestro A Coruña (España) ilarrosa@mundo-r.com
                Message 7 of 26 , Apr 3, 2007
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                  Francois Rideau wrote:
                  > Dear friends
                  > Here another construction of point P, (another, I hope for I don't
                  > read very carefully other posts!)
                  >
                  > 1° I construct the circle Gamma on A and B, tangent at A to line AC.
                  > 2° The perpendicular bissector of segment BC cuts Gamma in U and U'
                  > such that AU is the internal bissector of <BAC and AU' is the other
                  > (external) bissector of the same angle <BAC.

                  You meant the perpendicular bissector of segment AB, of course ...

                  Best regards,

                  Ignacio Larrosa Cañestro
                  A Coruña (España)
                  ilarrosa@...

                  > 3° The parallel to line AC through U cuts again Gamma in F.
                  > 4° Line FC cuts again circle Gamma in the sought point P
                  > Friendly
                  > Francois
                • Francois Rideau
                  Of course, yes! Sorry for mistaking a key on the keyboard, now I am feeling suddenly very old! Friendly Francois ... [Non-text portions of this message have
                  Message 8 of 26 , Apr 3, 2007
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                    Of course, yes! Sorry for mistaking a key on the keyboard, now I am feeling
                    suddenly very old!
                    Friendly
                    Francois

                    On 4/3/07, Ignacio Larrosa Cañestro <ilarrosa@...> wrote:
                    >
                    > Francois Rideau wrote:
                    > > Dear friends
                    > > Here another construction of point P, (another, I hope for I don't
                    > > read very carefully other posts!)
                    > >
                    > > 1° I construct the circle Gamma on A and B, tangent at A to line AC.
                    > > 2° The perpendicular bissector of segment BC cuts Gamma in U and U'
                    > > such that AU is the internal bissector of <BAC and AU' is the other
                    > > (external) bissector of the same angle <BAC.
                    >
                    > You meant the perpendicular bissector of segment AB, of course ...
                    >
                    > Best regards,
                    >
                    > Ignacio Larrosa Cañestro
                    > A Coruña (España)
                    > ilarrosa@... <ilarrosa%40mundo-r.com>
                    >
                    > > 3° The parallel to line AC through U cuts again Gamma in F.
                    > > 4° Line FC cuts again circle Gamma in the sought point P
                    > > Friendly
                    > > Francois
                    >
                    >


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