Something stronger: Prove that each set of lines which cover all the vertices of a board of side $n$ has at least $n + 1$ lines.

The answer is $n$. To cover the triangular board with only $n$ changes of direction it suffices to follow a spiral pattern, as illustrated in the figure below, for $n = 6$: [asy][asy] unitsize(20); pair h = (1, 0); pair v = dir(60); pair d = dir(120); for(int i = 0; i < 6; ++i) { draw(i*v -- i*v + (6 - i)*h); draw(i*h -- i*h + (6 - i)*v); draw((i + 1)*h -- (i + 1)*h + (i + 1)*d); } draw((0, 0) -- 6*h -- 6*v -- v -- v + 4*h -- h + 4*v -- h + 2*v -- 2*h + 2*v, linewidth(2)); draw(h + 2*v -- 2*h + 2*v, EndArrow); fill(circle((0, 0), 0.1)); [/asy][/asy] Now weshow that each path that covers all the vertices has at least $n$ changes of direction. First, let us observe that the number of changes of direction is one less than the number of segments that form the path. To prove the result, we will show that each set of lines which covers all the vertices of a triangular board of size $n$ has at least $n + 1$ of them. This is sufficient, since the lines containing the segments of the path have this property. The proof of this result proceeds by induction on $n$, where the base case $n = 1$ is easy to verify. Consider now an equilateral board of size $n$ with $n \geq 1$, let us focus on one of its sides. This side contains $n + 1$ vertices of the board. If each line covers at most one vertex of this side then clearly we use at least $n + 1$ of them. If on the other hand, a line passes through at least two of these vertices, then it passes through all, and so no other vertex of the board is covered by it . This means that the lines left must cover the vertices left, but this vertices correspond to a triangular board of size $n - 1$, and so by the induction hipothesis we need at least $n$ apart from the first one. Thus we have at least $n + 1$ lines in total. This completes the induction and the problem. Curiosity The vertices of a triangular board of size $n$ form the least (in terms of cardinality) set of points so that we cannot cover it with $n$ lines, but we can if we remove any of the points.