Inside Job

$AD=\sqrt{BD^2-AB^2}=\sqrt{26^2-10^2}=24$

Area $[ABD]=\frac12 AB\times AD=\frac12 \times 10\times 24=120$

Area $[BCD]=[ABCD]-[ABD]=306-120=186$

$AB$ is perpendicular to $AD$ and $EC$ is parallel to $AB$ , so $EC$ is perpendicular to $AD$ .

$[BCD]=[BCE]+[CED]=\frac12 EC h_1+\frac12 ECh_2=$

$\frac12 EC(h_1+h_2)=\frac12 EC\times AD=\frac12 24EC=12 EC$

$12EC=186$

$EC=\dfrac{186}{12}=\boxed{15.5}$