OMCPLP 2014 Problem 6

Kilua and Ndoti play the following game in a square $ABCD$ : Kilua chooses one of the sides of the square and draws a point $X$ at this side. Ndoti chooses one of the other three sides and draws a point Y. Kilua chooses another side that hasn't been chosen and draws a point Z. Finally, Ndoti chooses the last side that hasn't been chosen yet and draws a point W. Each one of the players can draw his point at a vertex of $ABCD$ , but they have to choose the side of the square that is going to be used to do that. For example, if Kilua chooses $AB$ , he can draws $X$ at the point $B$ and it doesn't impede Ndoti of choosing $BC$ . A vertex cannot de chosen twice. Kilua wins if the area of the convex quadrilateral formed by $X$ , $Y$ , $Z$ , and $W$ is greater or equal than a half of the area of $ABCD$ . Otherwise, Ndoti wins. Which player has a winning strategy? How can he play?