Square, square, and square

Let $a = |AB|$ and $b = |DE|$ . Then by the formula $\arctan(x) + \arctan(y) = \arctan\left(\dfrac{x + y}{1 - xy}\right)$ , we have

$\angle FJH = \angle GAF + \angle BHA = \arctan\left(\dfrac{b}{a + b}\right) + \arctan\left(\dfrac{a}{a + 2b}\right) =$

$\arctan\left(\dfrac{\dfrac{b}{a + b} + \dfrac{a}{a + 2b}}{1 - \dfrac{b}{a + b} \times \dfrac{a}{a + 2b}}\right) = \arctan\left(\dfrac{b(a + 2b) + a(a + b)}{(a + b)(a + 2b) - ab}\right) =$

$\arctan\left(\dfrac{2ab + 2b^{2} + a^{2}}{a^{2} + 2b^{2} + 2ab}\right) = \arctan(1) = 45^{\circ}$ , and thus $\angle AJH = 180^{\circ} - \angle FJH = \boxed{135^{\circ}}$ .

Let a = length of AB, and b = length of DE.

AB = BC = CD = DA = a, DE = EF = FG = GD = GH = HI = IF = b.

Let K is the point at AH so that JK perpendicular to AH, and extension of HI and extension of AF intersect at point L.

With analytic, we can know the length of GJ is equal to b. If we draw a circle centered at G with radius b, then point F, H, and J are on circle. Therefore, the angle of FJH is equal to half of the angle of FGH. We know that the angle of FGH is 90 degree. So, the angle of FJH is 45 degree. At last, the angle of AJH is 180 - 45 = 135 degree.

Note : If you got another solution without analytic, please post your solution. I will be glad to see your solution.

Let AD=y, EF=x. <AJH = <ABJ + < JAB.
tan(<AJH) = tan <ABJ + < JAB. = ((2x+y)/y +(x+y)/x )/(1- ( (2x+y)/y )((x+y)/x)) = (2x^2+2xy+y^2)/(-(2x^2+2xy+y^2)) = -1, <AJH =135.

Draw a perpendicular from J to AH meeting the latter in K and let DK = x, JK = y with AD = a & DG = b. We need not use analytical geometry but just similar triangles and Pythagoras to claim that : (2b-x)/(a+2b)=y/a,y/(a+x)=b/(a+b), z² =(b-x)²+y² where z = GJ. These yield: x =ab²/[(a+b)²+b²)], y={ab(a+2b)}/[(a+b)²+b²)]. More interestingly, z = b = GJ = GD = GF = GH or that G is the circum-centre of the circle passing through * D, J, F & H . Then as b4, the chord FH which subtends an angle of 90° at G must subtend an angle of 45° at J or angle * AJH=135°. Thanks for a nice solution, * Mr. Am Nu * and my apologies for not using LaTeX properrly

Let the side length of the big square be $3$ and the small square be $2$ .

$tan~\theta=\dfrac{2}{5}$ $\implies$ $\theta=21.80140949^\circ$

$tan~\phi=\dfrac{3}{7}$ $\implies$ $\phi=23.19859051^\circ$

Therefore,

$x=180-\theta-\phi=180-21.80140949-23.19859051=135^\circ$