MathLabs

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 4 of 6: Transfer the cyclic angles
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∠AEM=∠AEK=90∘,A,E,D,M are concyclic,∠AMB=∠AEF=∠ACF.\angle AEM=\angle AEK=90^\circ,\quad A,E,D,M\text{ are concyclic},\quad \angle AMB=\angle AEF=\angle ACF.
Detailed analysis

Because E lies on the ray MH and K is on the same line, EK is perpendicular to AE. Also AD is perpendicular to DM, so A,E,D,M are concyclic. Since E,D,F are collinear and A,B,C,F are on Gamma, the equal subtended angles give angle AMB=angle AEF=angle ACF.