hi Agnishom and bobbym
Here's two new diagrams.
I constructed the square ABCD.
I put point G somewhere on AD
I made an isosceles triangle FGC so that FG = FC
I found the centre of the square and rotated point F by 270 degrees around this point to fix H.
That makes BH = FG = FC
I constructed a line, perpendicular to GF at G and bisected the angle to make FGJ = 45.
As you can see in the first diagram J and H are different points.
But, this method of construction allows me to move G along the line. My plan was to find the position where J and H coincide.
Because of the size of the points, it's hard to be exact about this (Euclid had dimensionless points but they cost extra )
So I measured GHB and GJB and tried to make them equal. The second diagram shows my best attempt.
So the 90 case seems to be a special case and the only one where FGH = 45.
(Not counting G at A or B)
Your original diagram does make it look like there's a right angle there, but, in geometry, it is dangerous assuming things just because they look like it.
Later I'll have a go at the proof when I assume the angle is 90.