Length of the dividing line

In $\triangle ABC$ , $AD=12$ , $BD=5$ and $DC=9$ .

Since $FE \parallel AD$ , $\triangle FEC \sim \triangle ADC$ , and $\dfrac{FE}{EC}=\dfrac{AD}{DC}=\dfrac{4}{3}$

It follows that $EC=\dfrac{3}{4}FE$ .

Now the area of $\triangle ABC=\dfrac{1}{2}(14)(12)=84$ . The area of $\triangle FEC$ is $\dfrac{1}{2}(84)=42$ .

Therefore, the area of right $\triangle FEC=\dfrac{1}{2}(EC)(FE)=42$ . Substituting we get

$42=\dfrac{1}{2}(FE)(\dfrac{3}{4})(FE)$ $\implies$ $FE=4\sqrt{7}$

Finally,

$y=7$

Area of the $\triangle ABC=\sqrt{21(21-13)(21-14)(21-15)}=84$ , so $A_2=\dfrac{84}{2}=42$

$\cos C=\dfrac{14^2+15^2-13^2}{2\times14\times15}=\dfrac35, \sin C=\dfrac45$

If $\overline{CD}=a$ then $A_2=[DCE]=\dfrac12a\cos C a\sin C=\dfrac12\times\dfrac35\times\dfrac45\times a^2$

$a^2=42\times \dfrac{2\times25}{3\times4}=175, a=5\sqrt{7}$

$\overline{DE}=a\sin C=5\sqrt{7}\times\dfrac45=4\sqrt{7}$

Answer $=\boxed{7}$