Three parallel lines, l1, l2, l3 are intersected by another line l at points A, B, C respectively. Points E and D are taken onto l1, l3 so that BE=CB, BD=BA. Prove that midpoint K of the bisector BL of the angle DBE is equally apart of l1 and l3.
Proof:
Let l2 intersect DE at M, then EM/DM=AB/BC ( 1 ). Apply angle bisector theorem in tr. BDE, DL/EL=BD/BE=AB/BC ( 2 ), that is DL=EM, so N, midpoint of ML is midpoint of DE as well, so it lies at equal distance from l1 and l3. NK is midline in tr. BML, thus we are done.
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