Find The Length

First let's calculate the area of $\triangle ABC$ using Heron's formula: $\Delta=\sqrt{\dfrac{(6+8+4)(6+8-4)(6-8+4)(-6+8+4)}{4}}=\dfrac{21 \sqrt{15}}{4}$ Find the length of the heights $PB$ and $QC$ using the formula of the area of triangle knowing base and height:

$\Delta=\dfrac{8PB}{2} \Rightarrow PB=\dfrac{21 \sqrt{15}}{16}$

$\Delta=\dfrac{6QC}{2} \Rightarrow QC=\dfrac{7 \sqrt{15}}{4}$

Using Pythagorean theorem on $\triangle BPC$ and $\triangle BQC$ , find $PC$ and $BQ$ respectively:

$PB^2+PC^2=BC^2 \Rightarrow PC=\dfrac{77}{16}$

$QC^2+BQ^2=BC^2 \Rightarrow BQ=\dfrac{7}{4}$

Find $AQ$ and $AP$ :

$AQ=AB-BQ=\dfrac{17}{4}$

$AP=AC-PC=\dfrac{51}{16}$

Find $\cos \angle BAC$ using cosines law:

$\cos \angle BAC=\dfrac{AB^2+AC^2-BC^2}{2 \times AB \times AC}=\dfrac{17}{32}$

Find $PQ$ using again cosines law:

$PQ^2=AQ^2+AP^2-2AQ \times AP \times \cos \angle BAC \Rightarrow PQ=\dfrac{119}{32}$

Find $\cos \angle BPQ$ :

$\cos \angle BPQ=\dfrac{PQ^2+PB^2-BQ^2}{2\times PQ \times PB}=\dfrac{\sqrt{15}}{4}$

Now, move to the right triangle $\triangle PCB$ and find $\cos \angle PCB$ :

$\cos \angle PCB=\dfrac{PC}{BC}=\dfrac{11}{16}$

Now, move to the triangle $\triangle PXC$ :

We know that $\angle PXC + \angle PCB + \angle BPQ + 90° = 180°$ , so: $\cos \angle PXC = \cos(90°-\angle PCB-\angle BPQ)=\sin(\angle PCB+\angle BPQ)$

Use the Pythagorean identity and the double angle sum to find that $\sin \angle PXC=\dfrac{7}{8}$

Use sines law to find $BX$ :

$\dfrac{PC}{\sin \angle PXC}=\dfrac{BX+BC}{\sin(90°+\angle BPQ)}=\dfrac{BX+BC}{\cos(\angle BPQ)} \Rightarrow BX=\dfrac{21}{8}$

Find $\cos \angle ABC$ :

$\cos \angle ABC=\dfrac{AB^2+BC^2-AC^2}{2 \times AB \times BC}=\dfrac{1}{4}$

Now, move to the triangle $\triangle ABX$ and find $AX$ :

$AX^2=AB^2+BX^2-2\times AB \times BX \times \cos(180° \angle ABC)=AB^2+BX^2+2\times AB \times BX \cos \angle ABC \Rightarrow AX=\dfrac{57}{8}$

By Power of a Point, we know that $XY \times AX=BX \times (BX+BC)$ , so $XY=\dfrac{539}{152}$

Now, move to the triangle $\triangle AXB$ and find $\cos \angle AXB$ :

$\cos \angle AXB=\dfrac{AX^2+BX^2-AB^2}{2 \times AX \times BX}=\dfrac{11}{19}$

Now, move to the right triangle $\triangle XYJ$ and find $BJ$ :

$\cos AXB=\dfrac{XY}{XJ}=\dfrac{XY}{BX+BJ} \Rightarrow BJ=\dfrac{7}{2}$

Finally, move to the triangle $\triangle ABJ$ and find $AJ^2$ :

$AJ^2=AB^2+BJ^2-2 \times AB \times BJ \times \cos \angle ABC$

$AJ^2=\dfrac{151}{4}$

Hence, $m=151$ , $n=4$ and $m+n=\boxed{155}$ .

Use Apollonius Theorem of triangles! It becomes a 3 liner!

Sorry for the late reply. Here's the background: Apollonius Theorem

${AB}^{2} + {AC}^{2} = 2({AJ}^{2} + {0.5BC}^{2})$

Therefore ${6}^{2} + {8}^{2} = 2\left({AJ}^{2} + {\left(\frac{7}{2}\right)}^{2}\right)$ .

Isolating ${AJ}^{2}$ yields $\frac{151}{4}$ .

151 + 4 = 155