Ukrajina 2008

Question

BB_1 is an altitude of the acute-angled scalene triangle ABC. Point D is placed on the side BC so that BAD = CBB_1. Segments AD and BB_1 intersect at point F. Line l is drawn through point B at right angle to the side AB. This line intersects line CF at point K. Prove that line DK intersects segment BF at a midpoint. FIGURE_0

Accepted Answer

Let line AD intersect a circle circumscribed about triangle ABC at point N. Then BCN = BAN = CBB_1. This implies that BB_1 CN (fig.2), and therefore ACN = 90^ AN is a diameter. Thus NBA = 90^, which implies that N,B,K are on the same line. As diagonals of trapezium CFBN intersect at point D, its sides CF and BN extended intersect at point K. The well-known properties of trapezium imply the rest of the proof. FIGURE_0 **Fig.2**