19. As shown in the figure, ABCD is a square wall with each side measuring 340 meters. A police officer and a thief start from opposite corners B and D, respectively, and move counterclockwise along ABCD to engage in a pursuit and escape activity. It is known that the police officer walks 85 meters per minute, and the thief walks 75 meters per minute. After some time, the thief appears in the police officer's line of sight for the first time. Since the thief has a rearview mirror, he realizes that the police officer has already appeared in his rearview mirror, so he changes his escape strategy and starts to run perpendicular to the wall (for example, in the figure, if the thief is at point E and sees the police officer in his rearview mirror, he will start to run along EF). Due to the thief changing his escape route, the police officer also changes the pursuit route (for example, in the figure, the police officer will pursue directly along CF, and eventually catches the thief at point F). How many minutes does it take for the police officer to catch the thief as quickly as possible?
【Analysis】For the police to see the thief, both must be on the same edge, with the maximum distance difference being 340 meters. Therefore, the police need to catch up by 340 meters, which takes $340 \div (85-75) = 34$ minutes. At this point, the police have run $34 \times 85 = 2890$ meters, and $2890 \div 340 = 8 \cdots 170$, which is exactly in the middle of an edge, so the police cannot see the thief yet. The police need to run an additional 170 meters to see the thief, which takes $170 \div 85 = 2$ minutes. At this time, the thief has run $36 \times 75 = 2700$ meters, and $2700 \div 340 = 7 \cdots 320$, so the thief is 320 meters away from the police.

As shown in the figure below, suppose the police catch the thief after $t$ minutes. By the Pythagorean theorem,
we have $(75 t)^{2} + 320^{2} = (85 t)^{2} \Rightarrow t = 8$
Therefore, it takes a total of 44 minutes.