A line segment from a vertex to the opposite side

Relevant wiki: Stewart's Theorem

Apply Law of Cosines twice

5^2 = 4^2 + 6^2 - 2 X 4 X 6 X cos B

cos B = 0.5625

BE + EA = 6 : 1.5 EA = 6 : EA = 4 so BE = 2

CE^2 =4^2 + 2^2 - 2 X 4 X 2 X cosB = 11

CE = sqrt(11)

Using Heron's formula ,the area of $\triangle ABC$ is given by:

$\begin{aligned} [ABC] & = \sqrt{s(s-a)(s-b)(s-c)} & \small \color{#3D99F6} \text{where }s = \frac {a+b+c+d}2 \\ & = \sqrt{\frac {15}2\left(\frac {15}2 - 4 \right)\left(\frac {15}2 - 5 \right)\left(\frac {15}2 - 6 \right)} \\ & = \sqrt{\frac {15}2\left(\frac 72\right)\left(\frac 52 \right)\left(\frac 32 \right)} \\ & = \frac {15\sqrt 7}4 \end{aligned}$

Let $CF$ the altitude from $C$ to $AB$ and its length $h$ and $EF=x$ . Then, we have:

$\begin{aligned} \frac 12 \times AB \times CF & = [ABC] \\ \frac {6h}2 & = \frac {15\sqrt 7}4 \\ \implies h & = \frac {5\sqrt 7}4 \end{aligned}$

By Pythagorean theorem , we have:

$\begin{aligned} BF^2 & = BC^2-CF^2 \\ (x+{\color{#3D99F6}2})^2 & = 4^2 - \left(\frac {5\sqrt 7}4\right)^2 = \frac {81}{16} & \small \color{#3D99F6} \text{Note that }BE = \frac {EA}2 = \frac {AB}3 = 2 \\ \implies x & = \frac 14 \end{aligned}$

By Pythagorean theorem, we have:

$\begin{aligned} CE^2 & = CF^2+EF^2 = \left(\frac {15\sqrt 7}4\right)^2 + \left(\frac 14\right)^2 = 11 \\ \implies CE & = \boxed{\sqrt{11}} \end{aligned}$