In $\triangle ABC$ AB=13,BC=14,CA=15 . D,E,F lie on BC ,CA,DE respectively . AD ,DE,AF perpendicular to BC,AC,BF. segment DF =M/N(coprime) .M+N? 
In $\triangle ABC$, $AB=13,BC=14,CA=15$. Points $D,E,F$ lie on $BC ,CA,DE$ respectively such that $AD ,DE,AF$ perpendicular to $BC,AC,BF$. Segment $DF$ =$M/N$(coprime). $M+N$ is equal to what ?

What I did -
used herons formula to find area then found $AD$ using it. Using Pythagoras, I found $DC$ as $9$ so $BD$ becomes $5$. Then since  $\triangle DBF\sim\triangle ABC$. Then using similarity got $\frac{AB}{DB}=\frac{BC}{FB}=\frac{AC}{DF}$.
So $13/5=15/DF$. So $DF=75/13$. So $M+N=88$.
Where am I going wrong?
Answer shows $21$


You already obtained $AD = 12, CD = 9, BD = 5$
In $\triangle ACD, CD:AD:AC = 3:4:5$
As $\angle ADE = \angle ACB$, $\triangle ADE \sim \triangle ACD$. So we obtain,
$DE = \dfrac{36}{5}, AE = \dfrac{48}{5}$
Also as $ABDF$ is cyclic, $\angle ABF = \angle ADE = \angle ACB \implies \triangle ABF \sim \triangle ACD$
So, $AF = \dfrac{52}{5}, EF^2 = AF^2 - AE^2 = 16$
Then $DF = \dfrac{36}{5} - 4 = \dfrac{16}{5} = \dfrac{m}{n}$
$ \therefore m + n = 21$
