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

Labels:
angle bisector,
circle,
midpoint,
tangent

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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.

ReplyDeletehttp://img18.imageshack.us/img18/8658/problem54.png

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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