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 3 of 6: Locate K on MH
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△BCK≅△CBH⟹BKCH is a parallelogram⟹H,M,K are collinear.\triangle BCK\cong\triangle CBH\Longrightarrow BKCH\text{ is a parallelogram}\Longrightarrow H,M,K\text{ are collinear}.
Detailed analysis

The two equal angle pairs make triangles BCK and CBH congruent. Hence BK is parallel and equal to CH, so BKCH is a parallelogram. Its diagonals BC and HK bisect one another; since M is the midpoint of BC, H,M,K are collinear.