Morning Square

First solution If we draw a parallel line to $\overline{AB}$ passing through $E$ and cutting $\overline{AD}$ at $H$ , we can find out that $\angle AEH = \alpha$ . Also, if we draw the line segment $\overline{GD}$ we can realize that $\bigtriangleup ADG \cong \bigtriangleup ABG \Rightarrow \overline{DG} \cong \overline{BG}$ and $\bigtriangleup CFG \cong \bigtriangleup CEG \Rightarrow \overline{FG} \cong \overline{EG}$ so we can say that $D$ , $E$ and $G$ are collinear points, so: $\angle CDE =\alpha \Rightarrow \angle DEH = \alpha \Rightarrow x=2\alpha$ Second solution As $\bigtriangleup ABE \cong \bigtriangleup BCF$ we have $\angle FBC =\alpha$ and quadrilateral $FHEC$ cyclic, being $H \cong \overline{AE} \cap \overline{BF}$ . As $FHEC$ is cyclic, $\angle FHC = 45^{o}$ , $\angle BFE = 45^{o}-\alpha$ and $\angle ACH =\alpha$ . As $\overline{AC} \perp \overline{FE}$ , $GHEL$ is a cyclic quadrilateral, being $L \cong \overline{AC} \cap \overline{FE}$ . As $GHEL$ is cyclic and $\bigtriangleup FLG \cong \bigtriangleup ELG$ we have: $\angle LHG =\angle LEG=\angle LFG=45-\alpha \Rightarrow \angle LHC =\alpha \Rightarrow \angle GLH =x=2\alpha$

Angle BEA = 90 - a = angle GFC = angle GEC

So it would be:

(90 - a) + (90 - a) + x = 180 (BEC is 180 degrees)

Manipulate we get: X - 2a = 0 --> x = 2a

First Solution: Second Solution:

Let point D is (0,0), and wlog, the sides of the square is 2.So F(1,0), C(2,0), E(2,1),B(2,2), and A(0,2). Eqs. of line AC : y=-x + 2, and BC : y=2x -2. So G(4/3,2/3). Look at ΔAGE, by using cos rule we get cos x=3/5, we know that cos α = 2/√5, So cos 2 α = 2 (cos (α))^2 – 1 = 3/5 = cos x, so x=2 α

[My proof is essentially the same as the original one by Hjalmar, but with a shortcut. I think it still works.]

Consider the line BF and an additional line ED. By symmetry alone, these must intersect on AC, thus they must intersect at G, and therefore E, G and D are collinear. Then if we connect E to the midpoint of AD, it is immediate that $x = 2\alpha$ .