Geometry Problem
Click the figure below to see the complete problem 547 about Triangle, Transversal, Four Circumcircles, Concurrency.
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Level: High School, SAT Prep, College geometry
Saturday, December 11, 2010
Problem 547: Triangle, Transversal, Four Circumcircles, Concurrency
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ReplyDeleteConnect GD,GE,GC, GF
G is the intersecting point of circle C1 and C2. We will prove that quadrilaterals GECF and ADGF are cyclic
1 m(BGC)=180- m(A) ABGC cyclic
m(BGE)=180-m(BDE) DBGE cyclic
Replace m(BDE)=m(A)+m(EFC) we will get m(EGC)=m(EFC) so GECF is cyclic
2 We have m(BGD)=m(BED)=m(CEF)=m(GCF)
So m(BGC)=m(DGF)=180-m(A) so ADGF cyclic
Peter Tran
If G is intersection point of C1 and C4 then
ReplyDeleteA + BGC = A + DGF or BGD = CGF ???
Let see ▲DBG = ▲CGF
GCF = DBG, see arcs on C1
GDB = GFC see arcs on C4
=> BGD = CGF
Denote <AFD by x
ReplyDeleteLet circles C1 and C2 intersect at G
Denote angles of the triangle ABC at the vertices A, B, C respectively by the same letters A, B, C.
From the figure, we note that
B,G,E,D (lying on C2) being concyclic,
<BGD = <BED = <CEF = C – x,
A,B,G,C (lying on C1)being concyclic,
<AGB = <ACB = C,
So <AGD = < AGB - <BGD = C – (C-x) = x = <AFD
Thus G lies on the circle C4 (through A, D, F)
(i.e) C1, C2, C4 pass through G
Next <CGF = <AGF - < AGC = <ADF - <ABC
= (Pi – A –x) – B = C – x = <ACE - <CFE = <CEF
So G lies on the circle C3 (through C, E, F)
Hence C1, C2, C3, C4 pass through G.
Proove that the centers of C_1, C_2, C_3 and C_4 lies in the same cirle
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