EFGH is a square. FG = MG and EK = JK. FGL, EMG, EKL and EHJ are straight lines. $\angle FEL$ is twice of $\angle FLE$.

1. Find the angle $\angle FGM$ using the property of the square's diagonal.

Since $EFGH$ is a square and $EG$ is its diagonal, it bisects the right angle at $G$.
$\angle FGE = \frac{90^\circ}{2} = 45^\circ$
Since $M$ lies on $EG$, $\angle FGM = 45^\circ$.

2. Use the isosceles triangle $FMG$ to find $\angle FMG$.

Given $FG = MG$, $\triangle FMG$ is an isosceles triangle.
$\angle FMG = \angle MFG$
$\angle FMG = \frac{180^\circ - 45^\circ}{2} = \frac{135^\circ}{2} = 67.5^\circ$

1. Find $\angle FEL$ using the right-angled triangle $EFL$

Since $EFGH$ is a square and $FGL$ is a straight line, $\angle EFL = 90^\circ$.
In $\triangle EFL$, $\angle FEL + \angle FLE = 180^\circ - 90^\circ = 90^\circ$.
Given $\angle FEL = 2\angle FLE$, we have:
$2\angle FLE + \angle FLE = 90^\circ$
$3\angle FLE = 90^\circ \implies \angle FLE = 30^\circ$
$\angle FEL = 2 \times 30^\circ = 60^\circ$

2. Find $\angle GEL$ by subtracting the diagonal angle

$EG$ is the diagonal of square $EFGH$, so $\angle FEG = 45^\circ$.
$\angle GEL = \angle FEL - \angle FEG$
$\angle GEL = 60^\circ - 45^\circ = 15^\circ$

1. Find $\angle LEJ$ using the right angle of the square.

Since $EFGH$ is a square and $E, H, J$ are collinear, $\angle FEJ = 90^\circ$.
From earlier parts, $\angle FEL = 60^\circ$.
$\angle LEJ = \angle FEJ - \angle FEL = 90^\circ - 60^\circ = 30^\circ$.

2. Use the isosceles triangle property to find $\angle KJE$.

In $\triangle EKJ$, we are given $EK = JK$, which means it is an isosceles triangle.
Therefore, the base angles are equal: $\angle KJE = \angle KEJ$.
Since $K$ lies on $EL$, $\angle KEJ = \angle LEJ = 30^\circ$.
Thus, $\angle KJE = 30^\circ$.