Monday, May 19, 2008

Elearn Geometry Problem 78

Click the figure below to see the problem statement.



Zoom complete Problem 78
Angles in a circle, Perpendicular lines, Congruence, Midpoint, Perpendicular bisector. Level: High School, SAT Prep, College geometry

8 comments:

  1. Hi my name is André, i dont speak english...To solve the problem 78 is needed any non-euclidean geometry? I have no solution to problem 78! Thanks!

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  2. Hi Andre,
    You only need Euclidean Geometry. Level: High School.
    Good luck!

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  3. I solved this question drawing two parallels lines...Just similarity! i would like to know what happen when A and B are the same point (AB=0), i have no geometric solution for this case!

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  4. Just similarity or maybe..... congruence.
    If AB = 0 (A,M,B are coincident points) the given condition OM perpendicular to AB changes to OM perpendicular to Line(2). The solution could be the same.

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  5. AO,BO cuts the circle at Y,Z
    AC*AD=AY*AO=BZ*BO=BE*BF---(1)
    AD cuts BF at X
    XC*XD=XE*XF---(2)
    line FCH cuts ⊿ABX, by Menelaus' Theorem
    XF/FB*BH/HA*AC/CX=-1---(3)
    line DEG cuts ⊿BAX, by Menelaus' Theorem
    XD/DA*AG/GB*BE/EX=-1---(4)
    after simplification, we get
    HA*AG=GB*BH
    and hence yields the result

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    Replies
    1. Formula 1 is not completely exact. I suggest:
      AO cuts the circle at P and Y, BO cuts the circle at Q and Z
      AC*AD=AP*AY=BQ*BZ=BE*BF

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  6. Mi solucion aqui:
    http://www.mediafire.com/view/h0hqirnhxkaxl2i/P14-metrigeo.doc

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  7. hi i am clarence i am in barrow and i need help what go's with 78

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