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
Monday, May 19, 2008
Elearn Geometry Problem 78
Subscribe to:
Post Comments (Atom)
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!
ReplyDeleteHi Andre,
ReplyDeleteYou only need Euclidean Geometry. Level: High School.
Good luck!
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!
ReplyDeleteJust similarity or maybe..... congruence.
ReplyDeleteIf 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.
AO,BO cuts the circle at Y,Z
ReplyDeleteAC*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
Formula 1 is not completely exact. I suggest:
DeleteAO cuts the circle at P and Y, BO cuts the circle at Q and Z
AC*AD=AP*AY=BQ*BZ=BE*BF
Mi solucion aqui:
ReplyDeletehttp://www.mediafire.com/view/h0hqirnhxkaxl2i/P14-metrigeo.doc
hi i am clarence i am in barrow and i need help what go's with 78
ReplyDelete