First set up a hall-of-mirrors diagram. The nature of the given triangle makes for an unambiguous tiling of the plane with unit squares upon which are overlaid diagonals. The diagonals alternate slope -1 and slope +1, in a checkerboard pattern of the original squares. They actually form a grid of squares themselves, tilted at 45° to the ordinary grid, but spaced at units of sqrt(2).

All the triangle vertices are on unit positions of the coordinate plane.

The slope is given in reduced form, that is 314 and 379 are relatively prime, so the first time a triangle vertex is hit will be at (379,314). We ask, what type of a square was the last that the beam going through before hitting this vertex--which way was its diagonal (the hypotenuse) going?

The first column of squares was centered at x=0.5. The first row of squares was centered at y=0.5. The square we're concerned with is centered at x=378.5, y=313.5 and is thus the same parity as the first column but the opposite parity of the first row. Its diagonal runs between upper right and lower left.

The slope is slightly lower than 1, so the beam is paralleling the hypotenuse (diagonal) just above it, to meet it at the upper right corner of the square, in the triangle that's the upper left half of the square. In this particular reflection, then, the acute angle being hit is the one clockwise from the right angle. But is this angle an ultimate reflection of the original angle clockwise from the right angle, or the one counterclockwise from the right angle? We need to count if it has been reflected an odd or an even number of times.

Each time the connecting line has gone through a vertical integral x value (378 times, from 1 to 378) it has been reflected.  Each time it has gone through a horizontal integral y value (313 times) it has been reflected. Each time it has passed through a hypotenuse it has also been reflected. How many times is that?

Remember that the hypotenuses form a grid tilted at 45° to the original grid, with spacing sqrt(2) times that of the original. Let's take point (1,0) in the original coordinate system as the origin for the rotated and magnified coordinate system. Consider the grid itself as rotated counterclockwise and magnified by a factor of sqrt(2)--the begin and end points then are, relative to the new coordinate system, rotated clockwise and drawn in (i.e., shrunk).

Point (0,0) in the old system is then (-.5,.5) in the new. Point (379,314) is (346,-32) in the new.

Conversion done by this Mintoris Basic program:

Degree

Input "x",x

input "y",y

x=x-1

r=Sqrt(x*x+y*y)

th=ATan(y/x)

if x<0 Then th=th+180

th=th-45

x=r*cos(th)/Sqrt(2)

y=r*sin(th)/Sqrt(2)

Print x,y

Grid lines for x=0 through x=345 have been gone through, making 346 of them. Grid lines for y=0 through y=-31 have been gone through, totaling 32.

Total reflections therefore have been 378 + 313 + 346 + 32, which has an odd total. Therefore the endpoint represents a reflection of the acute angle that's counterclockwise from the right angle in the actual triangle; that is, the point (1,0).