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

Labels:
circumcircle,
concurrent,
transversal,
triangle

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http://img821.imageshack.us/img821/1783/problem547.png

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

ReplyDelete