Problem 4
Let ABC be an acute triangle. Let D be the foot of the perpendicular from A to BC, M the midpoint of BC, and H the orthocenter of ABC. Let E be the intersection of the circumcircle Gamma of ABC with the ray MH, and let F be the other intersection of line ED with Gamma. Prove that BF/CF = AB/AC, where XY denotes the length of segment XY.
Step 1 of 6: Remove the symmetric case
Detailed analysis
When AB=AC, reflection symmetry about the altitude gives BF=CF, so the assertion is immediate. By symmetry assume AB>AC for the rest.