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Show that AB/AC=BF/FC

Source: APMO 2012 #4

April 2, 2012
geometrycircumcircletrigonometryparallelogramgeometric transformationreflectionAsymptote

Problem Statement

Let ABC ABC be an acute triangle. Denote by D D the foot of the perpendicular line drawn from the point A A to the side BC BC , by MM the midpoint of BC BC , and by H H the orthocenter of ABC ABC . Let E E be the point of intersection of the circumcircle Γ \Gamma of the triangle ABC ABC and the half line MH MH , and F F be the point of intersection (other than EE) of the line ED ED and the circle Γ \Gamma . Prove that BFCF=ABAC \tfrac{BF}{CF} = \tfrac{AB}{AC} must hold.
(Here we denote XYXY the length of the line segment XYXY.)
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