In a triangle $\triangle{ABC}$ $\measuredangle{BAC}=90^{\circ}$. $D$ is the midpoint of $BC$ and $E\in AC$. $|ED|=x+6$ and $AB=3x-4$. For condition $|AE|>|EC|$, find maximum integer value of $x$.
Here is a diagram for the question:
Let $|DC|=k$ and since $|DC|=|BC|$ we can conclude with angle transportations:
$$
|DC|=|BD|=|AD|=k
$$
Observe that triangle $\triangle{ADC}$ is an isosceles triangle so $x+6<k$. And with triangular inequality we have $3x-4<2k$. Seems like i can't get any useful data and i have no idea to solve for now. Hints and solutions are appreciated.
Let $F$ be the midpoint of $AC$. Note that $DF\perp AC$ and
$$|DF|=\frac{|BA|}{2}=\frac{3x-4}{2}$$
If $|AE|>|EC|$, then $E$ is a point on the line segment $FC$. So $|DE|>|DF|$.
\begin{align*} \frac{3x-4}{2}&<x+6\\ x&<16 \end{align*}
The maximum integer value of $x$ is $15$.