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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

Labels:
angle,
circle,
congruence,
line,
perpendicular

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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

Mi solucion aqui:

ReplyDeletehttp://www.mediafire.com/view/h0hqirnhxkaxl2i/P14-metrigeo.doc