Proposed Problem
Click the figure below to see the complete problem 338 about Triangle, Circumcircle, Inscribed Circle, Exterior angle bisector, Concyclic points.
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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
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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
Let the common tangent at E meet AC extended at K.
ReplyDeleteLet < CKE = 2x so that < FEK = < EFK = 90 - x and let < CEK = y = < CBE.
So < ACE = < ABE = 2x + y and < ABC = 2x + 2y, hence á = 90 - x – y ….(1)
Further < CEF = (90 – x) – y = á = < ABG from (1)
Therefore < EBG = < ABG + < ABE = < CEF + < ACE = < AFE proving that
GBFE is Concyclic
(Note that á = alpha)
Sumith Peiris
Moratuwa
Sri Lanka