Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 54Tangent Circles, angle bisector, midpoint. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
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.
http://img18.imageshack.us/img18/8658/problem54.pngConnect FH, AE, OAB and OFGDrawing OM perpendicular to CD. OM will intersect arc CD at midpoint M .We will prove that M,E,B and G, M, H are collinear1. Tri. EAB and MOB are isosceles with EA//MOm(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 collinear2. 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 CDPeter Tran
Thank you for giving me a chance to comment.