Ukrajina, 2008

Geometry Difficulty 5.1 AIME, harder Prove it Ukraine

$BB_1$ is an altitude of the acute-angled scalene triangle $ABC$ . Point $D$ is placed on the side $BC$ so that $\angle BAD = \angle CBB_1$ . 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.

Solution

Let line $AD$ intersect a circle circumscribed about triangle $\triangle ABC$ at point $N$ . Then $\angle BCN = \angle BAN = \angle CBB_1$ . This implies that $BB_1 \parallel CN$ (fig.2), and therefore $\angle ACN = 90^\circ \Rightarrow AN$ is a diameter. Thus $\angle NBA = 90^\circ$ , 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.

Fig.2

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