Proposed Problem
Click the figure below to see the complete problem 338 about Triangle, Circumcircle, Inscribed Circle, Exterior angle bisector, Concyclic points.
See more:
Complete Problem 338
Level: High School, SAT Prep, College geometry
Saturday, August 15, 2009
Problem 338. Triangle, Circumcircle, Inscribed Circle, Exterior angle bisector, Concyclic points
Subscribe to:
Post Comments (Atom)
http://img34.imageshack.us/img34/1809/problem338.png
ReplyDeleteConnect EA, EB, EDO, DE and EC
Draw a line from O perpendicular to AC. This line intersect circle O at point M.
1. Tri OME and DFE are isosceles with common side ODE and OM//DF
From this we can have m(OEM)=m(DEF) and 3 points E,F and M are collinear .
Since M is the midpoint of arc AC so EM become angle bisector of angle AEC
2. Note that m(AEC)= m(A) +m(C) and m(AEB)= m(C)
So m(BEF)=(m(A)-m(C))/2
3. We have m(BGA)=m(A)- alpha and 2.alpha=m(A)+m(C)
So m(BGA)= (m(A)-m(C))/2 = m(BEF) and quadrilateral GBFE is concyclic
Peter Tran