For convenience, let the bottom vertex of the square =(0,0)
The slope of one of the intersecting lines is 2...and its equation is y =2x
The slope of anoteher of the intersecting lines is -1/2.....and its equation is
y = -1/2x + 2
The x intersection of these lines is
2x = -1/2x + 2
(5/2)x = 2
x =4/5 = .8
And the y coordiante is t = (2)(4/5) = 8/5 = 1.6
We have a triangle with the points ( .8, 1.6) , (1, 2) and (2,2)
The (squared) sides of this triangle are
[ (.8- 1)^2 + ( 1.6 - 2)^2 ] = .2
[ (.8 - 2)^2 + (1.6 - 2)^2 ] = 1.6
And 1
We can use the Law of Cosines to find the angle
1 = .2 + 1.6 - 2 ( sqrt ( .2) sqrt ( 1.6) cos (angle)
1 = .2 + 1.6 - 2 sqrt ( .32) cos (angle)
[ 1 - ,2 -1.6 ] / [- 2sqrt (.32) ] = cos (angle)
arccos ( [ 1 - .2 - 1.6 ] / [- 2sqrt (.32) ] ) = 45° = angle