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

6 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!

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

    ReplyDelete
  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!

    ReplyDelete
  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.

    ReplyDelete
  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

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

    ReplyDelete