See complete Problem 54
Tangent Circles, angle bisector, midpoint. Level: High School, SAT Prep, College geometry
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Monday, May 19, 2008
Elearn Geometry Problem 54
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No problema 53, temos que BM é bissetriz do ângulo CBE. Como ângulo CBM = ângulo DBM, temos que arcoCM = arco MD. Fica provado que M é ponto médio de CD.
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ReplyDeleteConnect FH, AE, OAB and OFG
Drawing OM perpendicular to CD. OM will intersect arc CD at midpoint M .
We will prove that M,E,B and G, M, H are collinear
1. Tri. EAB and MOB are isosceles with EA//MO
m(MOB)=m(EAB) (corresponding angles)
m(OMB)=90-1/2m(MOB)
m(ABE)=90-1/2m(EAB)
so m(OMB)=m(ABE) and M, E, B are collinear
2. Similarly with tri. MOG and HFG
With the same logic as step 1 we will prove that G,H,M are collinear.
So GH and BE meet at M , midpoint of arc CD
Peter Tran
Thank you for giving me a chance to comment.
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