The optimal solution seems very complicated.
One should use repeated bisections.
The first bisection is obvious, symmetry suggest connecting opposing inner corners.
The second bisection is should use one of the inner corners, and somehow divide the half star in two parts that take equal time to finish.
It should not be a single straight line; since the remaining inner corner is attractive to use, it should 'bend optimally' where the third bisection (separating A and B in the picture) connects.
The optimal solution of the first 3 bisections may look close to the left picture
However if we look at bisection 4 in area A and B, line 4a will probably connect to 3a, while 4b will not, bending 3a will have quadratic cost while saving linearly for area A, so bending a (very) little will be beneficial - if the other lines are shifted to spread the benefit to the other areas.
This effect can be extended to later bisections; meaning lines 2,3a,3b and further all probably will have an infinite number of bends in the optimal solution.
The second right picture shows a far from optimal solution:
2+sqrt(3) time can be used to trisect the area and contain the angel within an area within a 1.5 by sqrt(3)/2 box,
repeatedly halving the box on the long side will cost an additional sqrt(3) +1.5 time. Total: 6.964
The left most does one optimization: First bisecting C from A+B allows B to be bigger; I used the same directions, and made the time for A and C the same. Again , further steps (sub optimally) just half a bounding rectangle to get to the mentioned result; e.g. PQRS for region A
Just pretend the green area is a rectangle with the length of 3/4 units, then repeatedly divide it into two "identical" parts using the smaller dimension unused in the last step.
The devil can trap the angel in some of the regions bordered by the lines (red first, maroon second, green third), teleport to the same side as the angel if needed and eventually rob the angel of any space to move.