Canadian Mathematical Olympiad

Solution:
We're given that $A B < A D$. Since $C Y$ bisects $\measuredangle B C D$, $B Y = Y D$, so $Y$ lies between $D$ and $A$ on the circle, as in the diagram above, and $D Y > Y A$, $D Y > A B$. Similar reasoning confirms that $X$ lies between $B$ and $C$ and $B X > X C$, $B X > C D$. So if $A B X C D Y$ has 4 equal sides, then it must be that $Y A = A B = X C = C D$.
Let $\measuredangle B A X = \measuredangle D A X = \alpha$ and let $\measuredangle B C Y = \measuredangle D C Y = \gamma$. Since $A B C D$ is cyclic, $\measuredangle A + \measuredangle C = 180^{\circ}$, which implies that $\alpha + \gamma = 90^{\circ}$. The fact that $Y A = A B = X C = C D$ means that the arc from $Y$ to $B$ (which is subtended by $\measuredangle Y C B$) is equal to the arc from $X$ to $D$ (which is subtended by $\measuredangle X A D$). Hence $\measuredangle Y C B = \measuredangle X A D$, so $\alpha = \gamma = 45^{\circ}$. Finally, $B D$ is subtended by $\measuredangle B A D = 2 \alpha = 90^{\circ}$. Therefore $B D$ is a diameter of the circle.

Solution:
We're given that $A B < A D$. Since $C Y$ bisects $\measuredangle B C D$, $B Y = Y D$, so $Y$ lies between $D$ and $A$ on the circle, as in the diagram above, and $D Y > Y A$, $D Y > A B$. Similar reasoning confirms that $X$ lies between $B$ and $C$ and $B X > X C$, $B X > C D$. So if $A B X C D Y$ has 4 equal sides, then it must be that $Y A = A B = X C = C D$. This implies that the arc from $Y$ to $B$ is equal to the arc from $X$ to $D$ and hence that $Y B = X D$. Since $\measuredangle B A X = \measuredangle X A D$, $B X = X D$ and since $\measuredangle D C Y = \measuredangle Y C B$, $D Y = Y B$. Therefore $B X D Y$ is a square and its diagonal, $B D$, must be a diameter of the circle.