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Cevian triangle of ABC is similar to ABC

Source: CentroAmerican 2004

December 10, 2010

geometryparallelogramgeometry proposed

Problem Statement

ABC

is a triangle, and

E

and

F

are points on the segments

BC

and

CA

respectively, such that

\frac{CE}{CB}+\frac{CF}{CA}=1

and

\angle CEF=\angle CAB

. Suppose that

M

is the midpoint of

EF

and

G

is the point of intersection between

CM

and

AB

. Prove that triangle

FEG

is similar to triangle

ABC

.

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