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[EMHL] Droz-Farny theorem again

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  • Milorad Stevanovic
    Dear friends, We have earlier the discussion about the proof of Droz-Farny theorem. Here is one proof of Droz-Farny theorem. Theorem Let two orthogonal lines
    Message 1 of 1 , Jan 25, 2004
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      Dear friends,
      We have earlier the discussion about the proof
      of Droz-Farny theorem.

      Here is one proof of Droz-Farny theorem.
      Theorem
      Let two orthogonal lines p1,p2 pass through
      orthocenter H and intersect BC,CA,AB at points
      M1,N1,P1 and M2,N2,P2 ,then the midpoints
      M,N,P of segments M1M2,N1N2,P1P2 are
      collinear.
      Proof.
      Let we have the points in following order
      B,M1,H1,C,M2 and C,N2,N1,A and P1,A,H3,P2,B
      where H1 and H3 are the feet of altitudes AH and CH.
      The points M,N,P are collinear if and only if exist
      k such that

      v(HP)=k*v(HM)+(1-k)*v(HN) (1)

      where v(ST) denotes vector ST.
      Since M,N,P are the midpoints of segments
      M1M2,N1N2,P1P2 we have

      2v(HM)=v(HM1)+v(HM2),
      2v(HN)=v(HN1)+v(HN2), (2)
      2v(HP)=v(HP1)+v(HP2).

      From (1) and (2) we get

      v(HP1)+v(HP2)=k*[v(HM1)+v(HM2)]+
      (1-k)*[v(HN1)+v(HN2)]=
      k*[-(M1H/HP1)v(HP1)-(M2H/HP2)v(HP2)]+
      (1-k)*[(HN1/HP1)v(HP1)-(N2H/HP2)v(HP2)].

      Linear independency of vectors HP1 and HP2
      leads us to two conditions

      1+k*(M1H/HP1)-(1-k)*(HN1/HP1)=0
      1+k*(M2H/HP2)+(1-k)(N2H/HP2)=0

      These conditions are

      N1P1+k*M1N1=0
      N2P2+k*M2N2=0

      In these both conditions k would exist if it is
      true that

      M1N1/N1P1=M2N2/N2P2 (3)

      If we apply Menelaus theorem on
      triangle M1P1B and line CA and on
      triangle M2P2B and line CA we get

      M1N1/N1P1=(AB/P1A)*(M1C/BC) ,
      M2N2/N2P2=(AB/AP2)*(CM2/BC). (4)

      From (3) and (4) we have that the
      theorem would be proved if it is true that

      AP2/P1A=CM2/M1C (5)

      Let D be angle BP1H.Then we have

      AP2/P1H=(AH3+H3P2)/(P1H3-AH3)=
      (tanB+tanD)/(cotanD-tanB)=tan(B+D)*tanD

      CM2/M1C=(H1M2-H1C)/(M1H1+H1C)=
      (tan(B+D)-tanB)/(cotan(B+D)+tanB)=tan(B+D)*tanD.

      Theorem is proved.

      Best regards

      Milorad R.Stevanovic





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