Loading ...
Sorry, an error occurred while loading the content.
 

Re: [EMHL] Altitudes as diameters of circles

Expand Messages
  • Antreas P. Hatzipolakis
    ... Dear Paul, A general problem that Floor posted says that the circles with diameters three concurrent cevians have always as radical center the orthocenter
    Message 1 of 20 , Feb 1, 2002
      On Thursday, January 31, 2002, at 01:59 PM, Paul Yiu wrote:

      > Dear Antreas and friends,
      >
      > [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
      >
      > Draw the circles (Ka), (Kb), (Kc) having as diameters the
      > altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].
      >
      > *** Where is the radical center?
      >

      Dear Paul,

      A general problem that Floor posted says that the circles with
      diameters three concurrent cevians have always as radical center
      the orthocenter of ABC.

      Here are some other rad. center problems on the altitudes
      configuration:

      1. Which is the locus of the radical center of the circles
      (Ka, t*h_a), (Kb, t*h_b), (Kc, t*h_c) as t varies?
      [For t = 1/2 we have the case of the altitudes as diameters.]

      2. Which is the locus of the radical center of the circles
      (Ka, t*R), (Kb, t*R), (Kc, t*R) as t varies?

      3. Which is the locus of the radical center of the circles
      (Ka, t*a), (Kb, t*b), (Kc, t*c) as t varies?

      Antreas
    • Antreas P. Hatzipolakis
      On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis ... Let A A be the common chord of (Kb), (Kc) [A near to A] B B be the common chord of
      Message 2 of 20 , Feb 1, 2002
        On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis
        wrote:

        > Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
        >
        > Draw the circles (Ka), (Kb), (Kc) having as diameters the
        > altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].

        Let
        A'A" be the common chord of (Kb), (Kc) [A' near to A]
        B'B" be the common chord of (Kc), (Ka) [B' near to B]
        C'C" be the common chord of (Ka), (Kb) [C' near to C]

        Are the triangles

        1. ABC, A'B'C'

        2. ABC, A"B"C"

        perspective?

        Antreas
      • Antreas P. Hatzipolakis
        ... In my figure, the triangles seems to be perspective, and the midpoint of the perspectors line segment is the Nine Point Circle center N. Proof ? Antreas
        Message 3 of 20 , Feb 1, 2002
          On Friday, February 1, 2002, at 02:21 PM, Antreas P. Hatzipolakis wrote:

          >
          > On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis
          > wrote:
          >
          >> Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
          >>
          >> Draw the circles (Ka), (Kb), (Kc) having as diameters the
          >> altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].
          >
          > Let
          > A'A" be the common chord of (Kb), (Kc) [A' near to A]
          > B'B" be the common chord of (Kc), (Ka) [B' near to B]
          > C'C" be the common chord of (Ka), (Kb) [C' near to C]
          >
          > Are the triangles
          >
          > 1. ABC, A'B'C'
          >
          > 2. ABC, A"B"C"
          >
          > perspective?
          >

          In my figure, the triangles seems to be perspective, and the midpoint
          of the perspectors' line segment is the Nine Point Circle center N.
          Proof ?

          Antreas
        • Boutte Gilles
          Dear Antreas and all hyacinthists, ... In both case, there are not perspective (a = 11 , b = 10.50 , c = 6). Best regards, Gilles
          Message 4 of 20 , Feb 1, 2002
            Dear Antreas and all hyacinthists,


            > [APH]
            >
            > > Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
            > >
            > > Draw the circles (Ka), (Kb), (Kc) having as diameters the
            > > altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].
            >
            > Let
            > A'A" be the common chord of (Kb), (Kc) [A' near to A]
            > B'B" be the common chord of (Kc), (Ka) [B' near to B]
            > C'C" be the common chord of (Ka), (Kb) [C' near to C]
            >
            > Are the triangles
            >
            > 1. ABC, A'B'C'

            >
            > 2. ABC, A"B"C"
            >
            > perspective?
            >
            > Antreas


            In both case, there are not perspective (a = 11 , b = 10.50 , c = 6).

            Best regards,

            Gilles
          • Paul Yiu
            Dear Antreas, [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes. Draw the circles (Ka), (Kb), (Kc) having as diameters the altitudes AHa, BHb,
            Message 5 of 20 , Feb 1, 2002
              Dear Antreas,

              [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
              Draw the circles (Ka), (Kb), (Kc) having as diameters the
              altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].

              [PY]: Where is the radical center?

              [APH]: A general problem that Floor posted says that the circles with
              diameters three concurrent cevians have always as radical center
              the orthocenter of ABC.

              *** Thank you for reminding me of this very interesting result. I remember
              it now.


              [APH]: Here are some other rad. center problems on the altitudes
              configuration:

              1. Which is the locus of the radical center of the circles
              (Ka, t*h_a), (Kb, t*h_b), (Kc, t*h_c) as t varies?
              [For t = 1/2 we have the case of the altitudes as diameters.]

              2. Which is the locus of the radical center of the circles
              (Ka, t*R), (Kb, t*R), (Kc, t*R) as t varies?

              For (2), it is a fixed point P.

              For (1), it is a line through the point P.

              For (3), another line through the same point P.



              3. Which is the locus of the radical center of the circles
              (Ka, t*a), (Kb, t*b), (Kc, t*c) as t varies?
              best regards
              sincerely
              Paul
            • yiuatfauedu
              Dear Antreas, [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes. Draw the circles (Ka), (Kb), (Kc) having as diameters the altitudes AHa, BHb,
              Message 6 of 20 , Feb 1, 2002
                Dear Antreas,

                [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
                Draw the circles (Ka), (Kb), (Kc) having as diameters the
                altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].

                Here are some other rad. center problems on the altitudes
                configuration:

                1. Which is the locus of the radical center of the circles
                (Ka, t*h_a), (Kb, t*h_b), (Kc, t*h_c) as t varies?
                [For t = 1/2 we have the case of the altitudes as diameters.]

                2. Which is the locus of the radical center of the circles
                (Ka, t*R), (Kb, t*R), (Kc, t*R) as t varies?

                For (2), it is a fixed point P.

                *** This is indeed the circumcenter of KaKbKc.

                Best regards
                Sincerely
              • yiuatfauedu
                Dear Antreas, [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes. Draw the circles (Ka), (Kb), (Kc) having as diameters the altitudes AHa, BHb,
                Message 7 of 20 , Feb 2, 2002
                  Dear Antreas,

                  [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
                  Draw the circles (Ka), (Kb), (Kc) having as diameters the
                  altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].

                  Here are some other rad. center problems on the altitudes
                  configuration:

                  1. Which is the locus of the radical center of the circles
                  (Ka, t*h_a), (Kb, t*h_b), (Kc, t*h_c) as t varies?
                  [For t = 1/2 we have the case of the altitudes as diameters.]

                  2. Which is the locus of the radical center of the circles
                  (Ka, t*R), (Kb, t*R), (Kc, t*R) as t varies?

                  For (2), it is a fixed point P.

                  [PY]: This is indeed the circumcenter of KaKbKc.

                  *** This is true if the orthocenter is replaced by an arbitrary
                  point. If circles of equal radii are constructed with centers at the
                  midpoints of the cevians of P, then the radical axes are precisely
                  the perpendicular bisectors of the segments joining these midpoints,
                  and the radical center is the circumcenter of this midpoints of
                  cevians triangle.

                  Best regards
                  Sincerely
                  Paul
                • Antreas P. Hatzipolakis
                  ... Dear Paul, Here is another problem: Let A B C be the circumcevian triangle of H and (Ka), (Kb), (Kc) the circles with diameters AA , BB , CC
                  Message 8 of 20 , Feb 4, 2002
                    On Thursday, January 31, 2002, at 01:59 PM, Paul Yiu wrote:

                    > Dear Antreas and friends,
                    >
                    > [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
                    >
                    > Draw the circles (Ka), (Kb), (Kc) having as diameters the
                    > altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].
                    >
                    > *** Where is the radical center?
                    >

                    Dear Paul,

                    Here is another problem:
                    Let A'B'C' be the circumcevian triangle of H and
                    (Ka), (Kb), (Kc) the circles with diameters AA', BB', CC'
                    respectively.

                    Which is the radical center?

                    The same question in general: A'B'C' = the circumcevian triangle
                    of a point P.


                    Antreas
                  • Antreas P. Hatzipolakis
                    ... Let me rephrase the problem in this way (in order to compose a variation): Let ABC be a triangle, and AHa, BHb, CHc its altitudes meeting at H. Let A , B ,
                    Message 9 of 20 , Feb 4, 2002
                      On Monday, February 4, 2002, at 06:10 AM, Antreas P. Hatzipolakis wrote:

                      > Let A'B'C' be the circumcevian triangle of H and
                      > (Ka), (Kb), (Kc) the circles with diameters AA', BB', CC'
                      > respectively.
                      >
                      > Which is the radical center?
                      >

                      Let me rephrase the problem in this way (in order to compose
                      a variation):

                      Let ABC be a triangle, and AHa, BHb, CHc its altitudes meeting at H.

                      Let A', B', C' be the reflections of H in Ha, Hb, Hc, resp.
                      [now AA', BB', CC' are the circumcevians of H]

                      Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
                      resp..

                      Which is the radical center?

                      VARIATION:

                      Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
                      but in A,B,C resp.

                      Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
                      resp..

                      Which is now the radical center of the circles?

                      Antreas
                    • Paul Yiu
                      Dear Antreas, [APH]: Let A , B , C be the reflections of H not in Ha, Hb, Hc, resp., but in A,B,C resp. Let (Ka), (Kb), (Kc) be the circles with diameters
                      Message 10 of 20 , Feb 4, 2002
                        Dear Antreas,

                        [APH]: Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
                        but in A,B,C resp.

                        Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
                        resp..

                        Which is now the radical center of the circles?

                        *** These circles all pass through H, which is therefore the radical center.
                        This applies to an arbitrary point in place of H.

                        Best regards
                        Sincerely
                        Paul
                      • Antreas P. Hatzipolakis
                        ... I meant to say: with diameters HaA , HbB , HcC [anyway, and the above one is another variation!] Antreas
                        Message 11 of 20 , Feb 4, 2002
                          On Monday, February 4, 2002, at 02:43 PM, Antreas P. Hatzipolakis wrote:

                          >
                          > On Monday, February 4, 2002, at 06:10 AM, Antreas P. Hatzipolakis wrote:
                          >
                          >> Let A'B'C' be the circumcevian triangle of H and
                          >> (Ka), (Kb), (Kc) the circles with diameters AA', BB', CC'
                          >> respectively.
                          >>
                          >> Which is the radical center?
                          >>
                          >
                          > Let me rephrase the problem in this way (in order to compose
                          > a variation):
                          >
                          > Let ABC be a triangle, and AHa, BHb, CHc its altitudes meeting at H.
                          >
                          > Let A', B', C' be the reflections of H in Ha, Hb, Hc, resp.
                          > [now AA', BB', CC' are the circumcevians of H]
                          >
                          > Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
                          > resp..
                          >
                          > Which is the radical center?
                          >
                          > VARIATION:
                          >
                          > Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
                          > but in A,B,C resp.
                          >
                          > Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
                          > resp..
                          >

                          I meant to say: with diameters HaA', HbB', HcC'
                          [anyway, and the above one is another variation!]

                          Antreas

                          > Which is now the radical center of the circles?
                          >
                          > Antreas
                          >
                        • Antreas P. Hatzipolakis
                          ... Dear Paul, Yes. Sorry for not being careful! I sent a correction. Antreas
                          Message 12 of 20 , Feb 4, 2002
                            On Monday, February 4, 2002, at 02:52 PM, Paul Yiu wrote:

                            > Dear Antreas,
                            >
                            > [APH]: Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
                            > but in A,B,C resp.
                            >
                            > Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
                            > resp..
                            >
                            > Which is now the radical center of the circles?
                            >
                            > *** These circles all pass through H, which is therefore the radical
                            > center.
                            > This applies to an arbitrary point in place of H.
                            >

                            Dear Paul,

                            Yes. Sorry for not being careful!
                            I sent a correction.

                            Antreas
                          • Paul Yiu
                            Dear Antreas, [APH]: Let A , B , C be the reflections of H not in Ha, Hb, Hc, resp., but in A,B,C resp. Let (Ka), (Kb), (Kc) be the circles with diameters
                            Message 13 of 20 , Feb 4, 2002
                              Dear Antreas,

                              [APH]: Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
                              but in A,B,C resp.

                              Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
                              resp..


                              I meant to say: with diameters HaA', HbB', HcC'
                              ...
                              Which is now the radical center of the circles?

                              *** Still the same orthocenter!

                              Best regards
                              Sincerely
                              paul
                            • Antreas P. Hatzipolakis
                              ... Dear Paul, I am afraid that there is no way to go away from H ! :-) Let s try other variations: Let Ea, Eb, Ec be the midpoints of AH,BH,CH, resp. [ie
                              Message 14 of 20 , Feb 4, 2002
                                On Monday, February 4, 2002, at 03:08 PM, Paul Yiu wrote:

                                > Dear Antreas,
                                >
                                > [APH]: Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
                                > but in A,B,C resp.
                                >
                                > Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
                                > resp..
                                >
                                >
                                > I meant to say: with diameters HaA', HbB', HcC'
                                > ...
                                > Which is now the radical center of the circles?
                                >
                                > *** Still the same orthocenter!
                                >

                                Dear Paul,

                                I am afraid that there is no way to go away from H ! :-)

                                Let's try other variations:

                                Let Ea, Eb, Ec be the midpoints of AH,BH,CH, resp.
                                [ie EaEbEc is the Euler triangle of ABC]

                                Consider the circles with diameters EaHa, EbHb, ECHc.
                                Which is their radical center?

                                Still H !
                                [EaHa, EbHc, EcHc are chords of the nine point circle of ABC
                                intersecting at H. So, this is the very first variation
                                in this thread of problems]

                                Variation:

                                Let Ma, Mb, Mc be the midpoints HHa, HHb, HHc, resp.

                                Consider the circles with diameters AMa, BMb, CMc.

                                Which is their radical center?

                                Still H????

                                Antreas
                              • Antreas P. Hatzipolakis
                                ... Another Variation: Consider the circles with diameters EaMa, EbMb, EcMc [AHa = the a-altitude; Ea, Ma = the midpoints of AH, HHa, resp., etc] Which is
                                Message 15 of 20 , Feb 4, 2002
                                  On Monday, February 4, 2002, at 03:49 PM, Antreas P. Hatzipolakis wrote:

                                  >
                                  > On Monday, February 4, 2002, at 03:08 PM, Paul Yiu wrote:
                                  >
                                  >> Dear Antreas,
                                  >>
                                  >> [APH]: Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
                                  >> but in A,B,C resp.
                                  >>
                                  >> Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
                                  >> resp..
                                  >>
                                  >>
                                  >> I meant to say: with diameters HaA', HbB', HcC'
                                  >> ...
                                  >> Which is now the radical center of the circles?
                                  >>
                                  >> *** Still the same orthocenter!
                                  >>
                                  >
                                  > Dear Paul,
                                  >
                                  > I am afraid that there is no way to go away from H ! :-)
                                  >
                                  > Let's try other variations:
                                  >
                                  > Let Ea, Eb, Ec be the midpoints of AH,BH,CH, resp.
                                  > [ie EaEbEc is the Euler triangle of ABC]
                                  >
                                  > Consider the circles with diameters EaHa, EbHb, ECHc.
                                  > Which is their radical center?
                                  >
                                  > Still H !
                                  > [EaHa, EbHc, EcHc are chords of the nine point circle of ABC
                                  > intersecting at H. So, this is the very first variation
                                  > in this thread of problems]
                                  >
                                  > Variation:
                                  >
                                  > Let Ma, Mb, Mc be the midpoints of HHa, HHb, HHc, resp.
                                  >
                                  > Consider the circles with diameters AMa, BMb, CMc.
                                  >
                                  > Which is their radical center?
                                  >
                                  > Still H????
                                  >

                                  Another Variation:

                                  Consider the circles with diameters EaMa, EbMb, EcMc
                                  [AHa = the a-altitude; Ea, Ma = the midpoints of AH, HHa, resp., etc]

                                  Which is their radical center?


                                  Enough for today!

                                  Good night!

                                  Antreas
                                • Antreas P. Hatzipolakis
                                  On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis ... Perspector: (cotA ::) in barycentrics. Variation: (Ka),(Kb),(Kc) = the circles with
                                  Message 16 of 20 , Feb 6, 2002
                                    On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis
                                    wrote:

                                    > Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
                                    >
                                    > Draw the circles (Ka), (Kb), (Kc) having as diameters the
                                    > altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].
                                    >
                                    > A1 = (second tangent to (Ka) from B) /\ (second tangent to (Ka) from C)
                                    > [the first one is BC]
                                    >
                                    > B1 = (second tangent to (Kb) from C) /\ (second tangent to (Kb) from A)
                                    >
                                    > C1 = (second tangent to (Kc) from A) /\ (second tangent to (Kc) from B)
                                    >
                                    > The triangles ABC, A1B1C1 are perspective.

                                    Perspector: (cotA ::) in barycentrics.


                                    Variation:

                                    (Ka),(Kb),(Kc) = the circles with diameters (HHa), (HHb), (HHc), resp.

                                    (K'a) = the reflection of (Ka) in the perp. bisector of BC.
                                    (K'b) = the reflection of (Kb) in the perp. bisector of CA.
                                    (K'c) = the reflection of (Kc) in the perp. bisector of AB.


                                    A' = (2nd tangent to (K'a) from B) /\ (2nd tangent to (K'a) from C)
                                    [the first one is BC] [the first one is BC]

                                    B' = (2nd tangent to (K'b) from C) /\ (2nd tangent to (K'b) from A)

                                    C' = (2nd tangent to (K'c) from A) /\ (2nd tangent to (K'c) from B)

                                    The triangles ABC,A'B'C' are perspective.

                                    tanA
                                    Perspector: ( ----------- ::) in barycentrics
                                    5 - 4tan^2A


                                    Antreas
                                  • Antreas P. Hatzipolakis
                                    On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis ... Another variation: AHa, AHb, AHc = altitudes. Hab = Orth. Proj. of Ha on AB, Hac =
                                    Message 17 of 20 , Feb 7, 2002
                                      On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis
                                      wrote:

                                      > Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
                                      >
                                      > Draw the circles (Ka), (Kb), (Kc) having as diameters the
                                      > altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].
                                      >
                                      > A1 = (second tangent to (Ka) from B) /\ (second tangent to (Ka) from C)
                                      > [the first one is BC]
                                      >
                                      > B1 = (second tangent to (Kb) from C) /\ (second tangent to (Kb) from A)
                                      >
                                      > C1 = (second tangent to (Kc) from A) /\ (second tangent to (Kc) from B)
                                      >
                                      > The triangles ABC, A1B1C1 are perspective.
                                      >

                                      Another variation:

                                      AHa, AHb, AHc = altitudes.

                                      Hab = Orth. Proj. of Ha on AB,
                                      Hac = Orth. Proj. of Ha on AC.


                                      (Kab) = the circle with diameter HaHab
                                      (Kac) = the circle with diameter HaHac

                                      A'= (2nd tangent to (Kab) from B) /\ (2nd tangent to (Kac) from C)
                                      [the 1st tangent is AB] [the 1st tangent is AC]

                                      Similarly B', C'.

                                      A
                                      /\
                                      / \
                                      / \
                                      / \
                                      / \
                                      / \
                                      / \
                                      Hab \
                                      / Hac
                                      / Kab \
                                      / Kac \
                                      B-----------Ha----------C

                                      A'



                                      The triangles ABC, A'B'C' are perspective.

                                      1
                                      Perspector: ( ----------------- ::) = (tanAsin^2A ::) in Barycentrics.
                                      cotA*(1 + cot^2A)


                                      Antreas
                                    • Antreas P. Hatzipolakis
                                      On Thursday, February 7, 2002, at 02:00 PM, Antreas P. Hatzipolakis ... BHb, CHc ... Similarly define the circles (Kbc),(Kba);(Kca),(Kcb) H a = the 2nd
                                      Message 18 of 20 , Feb 7, 2002
                                        On Thursday, February 7, 2002, at 02:00 PM, Antreas P. Hatzipolakis
                                        wrote:

                                        > Another variation:
                                        >
                                        > AHa, AHb, AHc = altitudes.
                                        BHb, CHc
                                        >
                                        > Hab = Orth. Proj. of Ha on AB,
                                        > Hac = Orth. Proj. of Ha on AC.
                                        >
                                        >
                                        > (Kab) = the circle with diameter HaHab
                                        > (Kac) = the circle with diameter HaHac

                                        Similarly define the circles (Kbc),(Kba);(Kca),(Kcb)

                                        H'a = the 2nd intersection of (Kab),(Kac)
                                        [the first is Ha]
                                        that is, HaH'a is the common chord of (Kab),(Kac)

                                        H'b = the 2nd intersection of (Kbc),(Kba)
                                        [the first is Hb]

                                        H'c = the 2nd intersection of (Kca),(Kcb)
                                        [the first is Hc]

                                        Are the triangles

                                        1. HaHbHc, H'aH'bH'c

                                        2. ABC, H'aH'bH'c

                                        perspective?


                                        Antreas


                                        >
                                        > A'= (2nd tangent to (Kab) from B) /\ (2nd tangent to (Kac) from C)
                                        > [the 1st tangent is AB] [the 1st tangent is AC]
                                        >
                                        > Similarly B', C'.
                                        >
                                        > A
                                        > /\
                                        > / \
                                        > / \
                                        > / \
                                        > / \
                                        > / \
                                        > / \
                                        > Hab \
                                        > / Hac
                                        > / Kab \
                                        > / Kac \
                                        > B-----------Ha----------C
                                        >
                                        > A'
                                        >
                                        >
                                        >
                                        > The triangles ABC, A'B'C' are perspective.
                                        >
                                        > 1
                                        > Perspector: ( ----------------- ::) = (tanAsin^2A ::) in Barycentrics.
                                        > cotA*(1 + cot^2A)
                                        >
                                        >
                                        > Antreas
                                        >
                                      Your message has been successfully submitted and would be delivered to recipients shortly.