, 2004

Solution:

Fig. 1.
Observe that $KLMN$ is a parallelogram, $Q$ is the midpoint of $MK$ and hence $NL$ also passes through $Q$. Let $T$ be the point of intersection of $AC$ and $BD$; and let $S$ be the point of intersection of $BD$ and $MN$.

Consider the triangle $MNK$. Note that $SQ$ is parallel to $NK$ and $Q$ is the midpoint of $MK$. Hence $S$ is the midpoint of $MN$. Since $MN$ is parallel to $AC$, it follows that $T$ is the midpoint of $AC$. Now $Q$ is the circumcentre of $\triangle ABC$ and the median $BT$ passes through $Q$. Here there are two possibilities:
(i) $ABC$ is a right triangle with $\angle ABC = 90^\circ$ and $T = Q$; and
(ii) $T \neq Q$ in which case $BT$ is perpendicular to $AC$.

Suppose $\angle ABC = 90^\circ$ and $T = Q$. Observe that $Q$ is the circumcentre of the triangle $DCB$ and hence $\angle DCB = 90^\circ$. Similarly $\angle DAB = 90^\circ$. It follows that $\angle ADC = 90^\circ$ and $ABCD$ is a rectangle. This implies that $KLMN$ is a rhombus. Hence $LK / LM = 1$ and this gives $CD = CB$. Thus $ABCD$ is a square.

In the second case, observe that $BD$ is perpendicular to $AC$, $KL$ is parallel to $AC$ and $LM$ is parallel to $BD$. Hence it follows that $ML$ is perpendicular to $LK$. Similar reasoning shows that $KLMN$ is a rectangle.

Using $LK / LM = CD / CB$, we get that $CBD$ is similar to $LMK$. In particular, $\angle LMK = \angle CBD = \alpha$ say. Since $LM$ is parallel to $DB$, we also get $\angle BQK = \alpha$. Since $KLMN$ is a cyclic quadrilateral we also get $\angle LNK = \angle LMK = \alpha$. Using the fact that $BD$ is parallel to $NK$, we get $\angle LQB = \angle LNK = \alpha$. Since $BD$ bisects $\angle CBA$, we also have $\angle KBQ = \alpha$. Thus
$$QK = KB = BL = LQ$$
and $BL$ is parallel to $QK$. This gives $QM$ is parallel to $LC$ and
$$QM = QL = BL = LC$$
It follows that $QLCM$ is a parallelogram. But $\angle LCM = 90^\circ$. Hence $\angle MQL = 90^\circ$. This implies that $KLMN$ is a square. Also observe that $\angle LQK = 90^\circ$ and hence $\angle CBA = \angle LQK = 90^\circ$. This gives $\angle CDA = 90^\circ$ and hence $ABCD$ is a rectangle. Since $BA = BC$, it follows that $ABCD$ is a square.