The three points A, B, C form a triangle. AB=4, BC=5, AC=6. Let the angle bisector of \angle A intersect side BC at D. Let the foot of the perpendicular from B to the angle bisector of \angle A be E. Let the line through E parallel to AC meet BC at F. Compute DF.
Solution
Since AD bisects \angle A, by the angle bisector theorem \frac{AB}{BD}=\frac{AC}{CD}, so BD=2 and CD=3. Extend BE to hit AC at X. Since AE is the perpendicular bisector of BX, AX=4. Since B, E, X are collinear, applying Menelaus' Theorem to the triangle ADC, we have \frac{AE}{ED} \cdot \frac{DB}{BC} \cdot \frac{CX}{XA}=1. This implies that \frac{AE}{ED}=5, and since EF \parallel AC, \frac{DF}{DC}=\frac{DE}{DA}, so DF=\frac{DC}{6}=\frac{1}{2}.
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