MathDB

2

Part of 2011 Turkey MO (2nd round)

Problems(1)

Turkey NMO 2011 Problem 2

Source:

12/10/2011
Let ABCABC be a triangle D∈[BC]D\in[BC] (different than AA and BB).EE is the midpoint of [CD][CD]. F∈[AC]F\in[AC] such that FEC^=90\widehat{FEC}=90 and ∣AF∣.∣BC∣=∣AC∣.∣EC∣.|AF|.|BC|=|AC|.|EC|. Circumcircle of ADCADC intersect [AB][AB] at GG different than AA.Prove that tangent to circumcircle of AGFAGF at FF is touch circumcircle of BGEBGE too.
geometrycircumcircletrigonometrygeometric transformationreflectionperpendicular bisectorcyclic quadrilateral