Find length BC of triangle with incircle and circumcircle 
Some thoughts... some chord theorem to get the angle between PQ and BC... AB and AC are tangent to the circle, there has to be another theorem about that, perhaps ADE is isosceles and that helps looking at the angled PDE and AEQ, and with all this one should be able to figure the angle at A, and once I have that probably I can use yet another theorem on chords to determine the length BC from the length PQ...idk


Let $r$ and $R$ be the radii of the incircle and circumcircle, respectively. Then
$$\frac r{4R} =\sin \frac A2 \sin \frac B2 \sin \frac C2\tag1$$
Note that $ab= PD\cdot DQ =6$ and $ac= PE\cdot EQ =4$
$$r\cos \frac A2 =a \sin\frac A2 =\frac{DE}2=1$$
$$2R\sin A = BC = b+ c=\frac{6+4}{a}=10\sin\frac A2\tag2
$$
which leads to the ratio $\frac rR = \frac25$. Also
$$\tan \frac B2 = \frac r{b} = \frac{ra}6 = \frac1{3\sin A}, \>\>\>\>\>\>\>\tan \frac C2 = \frac1{2\sin A}
$$
Substitute above into (1)
$$\frac1{10} =\frac{\sin\frac A2}{\sqrt{(9\sin^2A+1)(4\sin^2A+1)}}
=\frac{\sqrt{\frac{1-\cos A}2}}{\sqrt{(10-9\cos^2 A)(5-4\cos^2A)}}\tag3
$$
which has the solution $\cos A=0$, or $\sin\frac A2 = \frac1{\sqrt2}$. Thus, per (2)
$$BC = 10\sin \frac A2 = 10\cdot \frac1{\sqrt2} = 5\sqrt2$$
Edit: as pointed out by @Intelligenti pauca, $\cos A= \frac56$ is also a valid solution to (3), which leads to
$BC = \frac5{\sqrt3}$.
