9. A piece of paper in the shape of a regular hexagon, as shown, is folded so that the three marked vertices meet at the centre $O$ of the hexagon. What is the shape of the figure that is formed?
A Six-pointed star
B Dodecagon
C Hexagon
D Square
E Equilateral Triangle
9. E Label vertices $A, B, C, D, E, E$ and $F$ as shown. Since the hexagon is regular, it can be divided into six equilateral triangles as shown. Therefore quadrilateral $O A B C$ is a rhombus and hence its diagonal $A C$ is a line of symmetry. Therefore, if vertex $B$ is folded onto $O$, the fold will be along $A C$. Similarly, if vertices $D$ and $F$ are folded onto $O$, the folds will be along $C E$ and $E A$ respectively. Hence the figure that is formed will be a triangle and, since all three of the rhombuses $O A B C, O C D E$ and $O E F A$ are made out of two congruent equilateral triangles, the lengths of their diagonals $A C, C E$ and $E A$ will be equal. Hence the shape $A C E$ that is formed is an equilateral triangle.