, 2011

Geometry Difficulty 6.7 National olympiad Prove it Croatia

Let $K$ and $L$ be points on a semicircle with diameter $\overline{AB}$ . Bisector of the side $\overline{AB}$ intersects the segment $\overline{KL}$ at a point $U$ so that $A$ and $K$ lie on one side of the bisector and $B$ and $L$ on the other. Let $N$ be the foot of perpendicular from the intersection of lines $AK$ and $BL$ to the line $AB$ , and let $V$ be a point on the line $KL$ such that $\angle VAU = \angle VBU$ . Prove that the lines $NV$ and $KL$ are perpendicular. (Iran TST 2009, modified)

Solution

Let $S$ be the center of the given semicircle, $C$ the intersection of the lines $AK$ and $BL$ , and $T$ the intersection of the lines $AB$ and $KL$ .
From the given condition it follows that the quadrilateral $ABUV$ is cyclic so by the power of a point theorem we have
$|TU| \cdot |TV| = |TA| \cdot |TB|.$

The quadrilateral $ABLK$ is also cyclic, so we have
$|TA| \cdot |TB| = |TK| \cdot |TL|.$

Since $\overline{AL}$ , $\overline{BK}$ and $\overline{CN}$ are altitudes of the triangle $ABC$ and $S$ is the midpoint of $AB$ , points $K$ , $L$ , $N$ and $S$ belong to the nine points circle of that triangle. Therefore the quadrilateral $NSLK$ is also cyclic, so
$|TU| \cdot |TV| = |TA| \cdot |TB| = |TK| \cdot |TL| = |TS| \cdot |TN|.$
Thereby the quadrilateral $NSUV$ is cyclic, and hence $\angle NVU = 180^\circ - \angle NSU = 90^\circ$ .

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