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Re: [EMHL] Fwd: quadrilateral

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  • Nikolaos Dergiades
    Dear Antreas and Alexey, [AZ] ... Here is a proof: If IA = a, IB = b, IC = c, ID = d angle x = (A+C)/2 and r is the radious of incircle ABCD then a =
    Message 1 of 9 , Jul 8, 2009
      Dear Antreas and Alexey,

      [AZ]
      > Dear colleagues!
      > Given circumscribed quadrilateral ABCD. I is its incenter,
      > M, N are
      > the midpoints of AC and BD.
      > Then IM/IN=AC/BD iff ABCD is cyclic.

      Here is a proof:
      If IA = a, IB = b, IC = c, ID = d
      angle x = (A+C)/2 and
      r is the radious of incircle ABCD then
      a = r/sin(A/2), b = r/sin(B/2)
      c = r/sin(C/2), d = r/sin(D/2)
      2.sin(A/2).sin(C/2) = cos(x-C) - cos(x)
      2.sin(B/2).sin(D/2) = cos(x) - cos(x+D)

      sin²(A/2)+sin²(C/2)=(1-cosA+1-cosC)/2
      =1-cos(x).cos(x-C)= P

      sin²(B/2)+sin²(D/2)=1-cos(x).cos(x+D) = Q

      4.IM² = 2(a²+c²)-AC²
      4.IN² = 2(b²+d²)-BD²
      Since
      AC² = a²+c²-2ac.cos(360-D-A/2-C/2) or
      AC² = a²+c²-2ac.cos(x+D)
      BD² = b²+d²-2bd.cos(360-C-B/2-D/2) or
      BD² = b²+d²+2bd.cos(x-C)
      applying ratio properties we get
      IM²/IN² = AC²/BD² <=>

      AC²/BD² = (a²+c²)/(b²+d²) <=>

      (a²+c²)/(b²+d²)= -ac.cos(x+D)/bd.cos(x-C) <=>

      [sin²(A/2)+sin²(C/2)]/[sin²(B/2)+sin²(D/2)]=
      =-sin(A/2)sin(C/2)cos(x+D)/sin(B/2)sin(D/2)cos(x-C) <=>

      [1-cos(x).cos(x-C)]/[1-cos(x).cos(x+D)]=
      [cos(x)-cos(x-C)]cos(x+D)/[cos(x)-cos(x+D)]cos(x-C) <=>
      P/Q = R/S = (P-R)/(Q-S) = 1 <=> P = Q (1)
      because
      P-R = Q-S = 1-cos(x)[cos(x-C)+cos(x+D)]+cos(x-C)cos(x+D)
      and P - R =/= 0 because
      P-R > 1-[cos(x-C)+cos(x+D)]+cos(x-C)cos)x+D) or
      P-R > [1-cos(x)][1-cos(x+D)] > 0
      Hence from (1) we get
      P = Q <=> 1-cos(x).cos(x-C) = 1-cos(x).cos(x+D) <=>
      cos(x)[cos(x-C)-cos(x+D)] = 0 <=>
      2cos(x).sin((C+D)/2).sin((A+D)/2) <=>
      cos(x) = 0 <=> (A+C)/2 = 90 <=> A+C = 180
      or ABCD is cyclic.

      Is there a simpler proof?

      Best regards
      Nikos Dergiades












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    • xpolakis
      [AZ] ... [ND] ... [trigonometric] ... How about a geometric proof (simpler or not)? APH
      Message 2 of 9 , Jul 8, 2009
        [AZ]
        > > Given circumscribed quadrilateral ABCD. I is its incenter,
        > > M, N are
        > > the midpoints of AC and BD.
        > > Then IM/IN=AC/BD iff ABCD is cyclic.

        [ND]
        > Here is a proof:
        [trigonometric]

        > Is there a simpler proof?

        How about a geometric proof (simpler or not)?

        APH
      • Andreas Hatzipolakis
        ... From: alexey_zaslavsky Date: Thu, Jul 9, 2009 at 8:26 AM Subject: Re: Fwd: quadrilateral To: Hyacinthos-owner@yahoogroups.com
        Message 3 of 9 , Jul 8, 2009
          ---------- Forwarded message ----------
          From: alexey_zaslavsky <alexey_zaslavsky@...>
          Date: Thu, Jul 9, 2009 at 8:26 AM
          Subject: Re: Fwd: quadrilateral
          To: Hyacinthos-owner@yahoogroups.com


          Dear Antreas and Nikolaos!

          I also found this trigonometric proof. Another proof isn't elementary.
          Let the bisectors of angles between the diagonals
          intersect the sides of ABCD in points P, Q, R, S. Then there exists an
          ellipse touching the sides in these points.
          The center of this ellipse divide MN in relation AC/BD. So in
          considered case this ellipse coincide with the incircle
          and the segments joining opposite touching points are perpendicular.
          This is true iff ABCD is cyclic.

          Sincerely                                    Alexey
          >
          > [ND]
          > > Here is a proof:
          > [trigonometric]
          >
          > > Is there a simpler proof?
          >
          > How about a geometric proof (simpler or not)?
          >
          > APH
          >
        • ndergiades@yahoo.gr
          Dear Antreas and Alexey, One hour ago I found a similar proof: LEMMA: If a line divides the connections AA , BB of two segments AB, A B in parts as AB : A B
          Message 4 of 9 , Jul 9, 2009
            Dear Antreas and Alexey,
            One hour ago I found a similar proof:

            LEMMA: If a line divides the connections
            AA', BB' of two segments AB, A'B'
            in parts as AB : A'B' then this line
            is parallel to the bisector of the
            angle of the segments AB, A'B'.
            PROOF:
            If the line meets AA' at A" and BB' at B"
            such that AA"/A"A' = BB"/B"B' = AB/A'B'
            then constructing the pararellograms
            A"ABC, A"A'B'C' the line A"B" is the
            bisector of triangle A"CC'.

            In our quadrilateral ABCD the points
            I, M, N are collinear (Newton's theorem).
            Divide by E,G the sides AB, CD such that
            AE/EB = CG/GD = AC/BD = AM/BN.
            By lemma the line EG passes through I
            and is parallel to one bisector of the diagonals.
            Similarly divide the sides BC, DA by F, H
            such that AH/HD = CF/FB = AC/BD = AM/BN.
            By lemma the line FH is parallel to the other
            bisector of the angle of the digonals
            is perpendicular to EG and passes
            through I. Hence FH is perpendicular bisector
            of EG since the EFGH is parallelogram.
            If K is the intersection of AB, CD then
            since in triangle KEG, KI is bisector and median
            KI is perpendicular bisector of EG and coinsides
            with HF. Similarly the angle bisector of AD, BC
            coinsides with EG. Hence ABCD is cyclic since
            the angle bisector of opposite sides are
            perpendicular.

            Best regards
            Nikos Dergiades


            >
            >
            > Dear Antreas and Nikolaos!
            >
            > I also found this trigonometric proof. Another proof isn't
            > elementary.
            > Let the bisectors of angles between the diagonals
            > intersect the sides of ABCD in points P, Q, R, S. Then
            > there exists an
            > ellipse touching the sides in these points.
            > The center of this ellipse divide MN in relation AC/BD. So
            > in
            > considered case this ellipse coincide with the incircle
            > and the segments joining opposite touching points are
            > perpendicular.
            > This is true iff ABCD is cyclic.
            >
            > Sincerely                                
            >    Alexey





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          • xpolakis
            [ND] ... Dear Nikos, Nice proof! By the way, do you know the Greek newspaper C. CARATHEODORY published by THE FRIENDS OF CARATHEODORY ? (Look at:
            Message 5 of 9 , Jul 10, 2009
              [ND]
              > One hour ago I found a similar proof:

              Dear Nikos,

              Nice proof!

              By the way, do you know the Greek newspaper
              C. CARATHEODORY published by "THE FRIENDS OF CARATHEODORY" ?

              (Look at: http://www.karatheodori.gr/ )

              Today I received issue #42 (May-June 2009).

              In page 6, I read this problem:

              Let P be a point inside parallelogram ABCD,
              such that angles ABP = ADP.
              To prove that angles PAB = PCB


              APH
            • Nikolaos Dergiades
              Dear Antreas and Alexey [APH] ... Thank you very much. I didn t know this knewspaper. [AZ] ... Very nice observation! But there is a more interesting result.
              Message 6 of 9 , Jul 10, 2009
                Dear Antreas and Alexey

                [APH]
                > By the way, do you know the Greek newspaper
                > C. CARATHEODORY published by "THE FRIENDS OF CARATHEODORY"
                > ?
                >
                > (Look at:  http://www.karatheodori.gr/ )
                >
                > Today I received issue #42 (May-June 2009).
                >
                > In page 6, I read this problem:
                >
                > Let P be a point inside parallelogram ABCD,
                > such that angles ABP = ADP.
                > To prove that angles PAB = PCB

                Thank you very much.
                I didn't know this knewspaper.

                [AZ]
                > Both conditions define an equilateral hyperbola passing
                > through A, B,
                > C, D. There is an equivalent problem in our with Akopjan
                > book.

                Very nice observation!

                But there is a more interesting result.
                If P is an arbitrary point then the reflexion D' of D
                in PA and the reflexion C' of C in PB are equidistant
                from the circumcenter O of PAB. i.e. OC' = OD'.
                (See the following lemma)
                Hence in case of /_ABP = /_ADP = /_AD'P the point
                D' lies on the circumcircle of PAB and by the lemma
                C' lies on the circumcircle and we get
                /_PAB = /_PC'B = /_PCB.

                LEMMA. If ABC is a triangle with circumcenter O and
                BB' = CC' are two vectors then the reflexion
                B" of B' in AB and the relexion C" of C' in AC
                are equidistant from O with angle B"OC" = 2.A

                I have a nice proof but I have no time to write it
                and since I am leaving home to Mytilene I will see
                your clever comments I hope after three days
                from an Internet cafe.

                Best regards
                Nikos










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              • Vladimir Dubrovsky
                Dear Antreas and Alexey and Nikos ... I think the shortest, at least the most elementary, solution is to notice that if Q is the translate of P by vector
                Message 7 of 9 , Jul 10, 2009
                  Dear Antreas and Alexey and Nikos

                  > [APH]
                  >> Let P be a point inside parallelogram ABCD,
                  >> such that angles ABP = ADP.
                  >> To prove that angles PAB = PCB

                  > [AZ]
                  >> Both conditions define an equilateral hyperbola passing
                  >> through A, B, C, D. There is an equivalent problem in our with Akopjan
                  >> book.

                  > [ND]
                  > If P is an arbitrary point then the reflexion D' of D
                  > in PA and the reflexion C' of C in PB are equidistant
                  > from the circumcenter O of PAB. i.e. OC' = OD'.
                  > (See the following lemma)
                  > Hence in case of  /_ABP = /_ADP = /_AD'P the point
                  > D' lies on the circumcircle of PAB and by the lemma
                  > C' lies on the circumcircle and we get
                  > /_PAB = /_PC'B = /_PCB.

                  I think the shortest, at least the most elementary, solution is to notice that
                  if Q is the translate of P by vector AB=CD, then both equations for angles
                  are equivalent to the cyclicity of the quad BPCQ (<BCQ=<ADP=<ABP=<BPQ
                  and similarly for the second equation).

                  > LEMMA. If ABC is a triangle with circumcenter O and
                  > BB' = CC' are two vectors then the reflexion
                  > B" of B' in AB and the relexion C" of C' in AC
                  > are equidistant from O with angle B"OC" = 2.A
                  >
                  > I have a nice proof but I have no time to write it
                  > and since I am leaving home to Mytilene I will see
                  > your clever comments I hope after three days
                  > from an Internet cafe.

                  Nikos, I wonder if your proof is based on the fact that the
                  composition of two reflections
                  is a rotation by twice the angle between their axes.

                  Best regards,
                  Vladimir
                • Nikolaos Dergiades
                  ... [VD] ... Yes. You are right. Very nice proof! [ND] ... [VD] ... Not exactly. If O is the circumcenter of ABC and Ob is the reflexion of O in AB and Oc is
                  Message 8 of 9 , Jul 14, 2009
                    > Dear Vladimir,

                    > > [APH]
                    > >> Let P be a point inside parallelogram ABCD,
                    > >> such that angles ABP = ADP.
                    > >> To prove that angles PAB = PCB
                    >
                    > > [ND]
                    > > If P is an arbitrary point then the reflexion D' of D
                    > > in PA and the reflexion C' of C in PB are equidistant
                    > > from the circumcenter O of PAB. i.e. OC' = OD'.
                    > > (See the following lemma)
                    > > Hence in case of  /_ABP = /_ADP = /_AD'P the point
                    > > D' lies on the circumcircle of PAB and by the lemma
                    > > C' lies on the circumcircle and we get
                    > > /_PAB = /_PC'B = /_PCB.
                    [VD]
                    > I think the shortest, at least the most elementary,
                    > solution is to notice that
                    > if Q is the translate of P by vector AB=CD, then both
                    > equations for angles
                    > are equivalent to the cyclicity of the quad BPCQ
                    > (<BCQ=<ADP=<ABP=<BPQ
                    > and similarly for the second equation).

                    Yes. You are right.
                    Very nice proof!

                    [ND]
                    > > LEMMA. If ABC is a triangle with circumcenter O and
                    > > BB' = CC' are two vectors then the reflexion
                    > > B" of B' in AB and the relexion C" of C' in AC
                    > > are equidistant from O with angle B"OC" = 2.A
                    > >
                    > > I have a nice proof but I have no time to write it
                    > > and since I am leaving home to Mytilene I will see
                    > > your clever comments I hope after three days
                    > > from an Internet cafe.
                    [VD]
                    > Nikos, I wonder if your proof is based on the fact that
                    > the
                    > composition of two reflections
                    > is a rotation by twice the angle between their axes.

                    Not exactly.

                    If O is the circumcenter of ABC and
                    Ob is the reflexion of O in AB and
                    Oc is the reflexion of O in AC then
                    since for vectors BB' = CC', BOb= OA = COc
                    the triangles BB'Ob, CC'Oc are congruent
                    and directly similar,
                    the triangles BB'Ob, BB"O are congruent
                    and inversely similar,
                    the triangles CC'Oc, CC"O are congruent
                    and inversely similar, we conclude that
                    the triangles BB"O, CC"O are congruent
                    and directly similar.
                    Since the rotation about O by an angle 2.A
                    moves B on C, B" must move on C".
                    Hence PB" = OC" and angle B"OC" = 2.A

                    Best regards
                    Nikos



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