[asy][asy]pair A=(0,sqrt(343)),B=(-7,0),C=(7,0),D=(A+C)/2,E=(B+D)/2; draw(D--A--B--C--D--B^^C--E); label("$A$",A,N);label("$B$",B,SW);label("$C$",C,SE);label("$E$",E,NW);label("$D$",D,NE);[/asy][/asy] Let $BD=x$. Since $E$ is the midpoint of $BD$, this means $DE=\frac12BD=\frac x2$. Also, since $D$ is the midpoint of $CA$, this means $CD=\frac12AC=7\sqrt2$. Hence, $$\frac{CD}{DE}=\frac{AB}{BC}\implies\frac{7\sqrt2}{\frac x2}=\frac{14\sqrt2}{BC}\implies BC=x$$Therefore, $BD=BC$, so $\triangle BDC$ is an isosceles triangle, hence, $\triangle ABC\sim\triangle BDC$. And finally, we have: $$\frac{BD}{DC}=\frac{AB}{BC}\implies\frac x{7\sqrt2}=\frac{14\sqrt2}x\implies x=\boxed{14}$$