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 5 of 6: Apply the two similarities
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△ABM∼△AFC⟹BMAM=FCAC,△ACM∼△AFB⟹CMAM=FBAB.\triangle ABM\sim\triangle AFC\Longrightarrow\frac{BM}{AM}=\frac{FC}{AC},\quad \triangle ACM\sim\triangle AFB\Longrightarrow\frac{CM}{AM}=\frac{FB}{AB}.
Detailed analysis

The angle relation above together with angle ABM=angle AFC gives ABM similar to AFC. Interchanging B and C gives ACM similar to AFB, yielding the two side ratios.