Solution:

Construct segments $K L, L M$ and $M K$. Next, construct the three altitudes of triangle $K L M$ that meet in its orthocenter, $O$. Since $K L$ connects the two midpoints of $A B$ and $B C$ we know that $K L \| A B$ it is easy to see that $M T\|K O\| L Q$ since those lines are perpendicular to a pair of parallel lines. Similarly, $K T\|M O\| L S$ and $K Q\|L O\| M S$.

Note that triangle $K L M$ divides the original triangle $A B C$ into four congruent parts, all similar to the original triangle (and thus all acute), so the perpendiculars we dropped from $K, L$, and $M$ are just altitudes of the smaller triangles. Since all the smaller triangles are acute, all the altitudes will meet inside the respective triangles.

Because of all the parallel lines, we know that $O L S M, O M T K$ and $O K Q L$ are all parallelograms having diagonals $M L, K M$, and $L K$, respectively. The diagonal of a parallelogram divides it into two congruent triangles, so triangle $L S M$ is congruent to triangle $M O L$, triangle $M T K$ is congruent to triangle $K O M$, and triangle $K Q L$ is congruent to triangle $L O K$. If we consider the hexagon $K Q L S M T$ to be composed of triangle $K L M$ and the three outer pieces, we can see that triangle $K L M$ is composed of three smaller triangles that are congruent to the corresponding outer pieces, so the area of the hexagon is twice the area of triangle $K L M$.

But triangle $K L M$ connects the midpoints of the edges of triangle $A B C$ so each of its sides is half the length of the corresponding side of triangle $A B C$, so triangle $K L M$ is similar to triangle $A B C$, but with $1 / 4$ the area. We previously showed that the area of $K Q L S M T$ is twice the area of triangle $K L M$, so the area of the hexagon is $1 / 2$ the area of triangle $A B C$.