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
 A: 
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
$BC = \frac5{\sqrt3}$.
A: Here are 2 solutions 1) an analytical solution 2) a solution inspired by the solution of @Quanto.

*

*First solution; Let us take coordinate axes as displayed in Fig. 1 below.

Line $PQ$ is taken as the $x$-axis, with origin and unit such that $P,D,E,Q$ have resp. abscissas $-3,-1,1,2$.
Triangle $ADE$ is isosceles, due to the fact that $A$ is the intersection point of tangents in $D$ and $E$ to the incircle ; therefore $A$ belongs to the ordinate axis with coordinates $(0,\tan a)$, where  $a$:=angle($ODA$).

Fig. 1.
As a consequence:
$$AD=AE=\dfrac{1}{\cos a} \ \implies \ \vec{AD}=\binom{-1}{-\tan a}  \ \text{and} \ \vec{AE}=\binom{1}{-\tan a}\tag{1}$$
A classical theorem on intersecting chords in a circle gives:
$$AD.DB=PD.PQ=6 \ \implies \ DB=\dfrac{6}{AD}=6 \cos a\tag{2}$$
For a similar reason:
$$EC=4 \cos a\tag{3}$$
Therefore, using (1) and (2) and the trigonometric formula $1+\tan^2 a=1/\cos^2 a$:
$$\vec{AB}=\dfrac{\dfrac{1}{\cos a}+6 \cos a}{\dfrac{1}{\cos a}} \vec{AD}=(1+6 \cos^2 a)\binom{-1}{-\tan a}$$
implying
$$B=\binom{-1-6 \cos^2 a}{- 6 \sin a\cos a}\tag{4}$$
For completely similar reasons, using (3):
$$C=\binom{1+4 \cos^2 a}{- 4\sin a \cos a}\tag{5}$$
As a consequence of (4) and (5):
$$\vec{BC}=\binom{2+10 \cos^2 a}{2 \cos a \sin a}\tag{6}$$
It remains to express that line segment $BC$ must be tangent to the incircle. This will be done by the following condition:
$$BC=BD+CE\tag{7}$$
Squaring (7) and using (2), (3) and (6):
$$\underbrace{(2+10 \cos^2 a)^2+(2 \cos a \sin a)^2}_{BC^2}
=\underbrace{(10 \cos a)^2}_{(BD+CE)^2}\tag{8}$$
Setting $x=\cos^2 a$ in (8), we get the following quadratic equation:
$$(1+5x)^2+x(1-x)=25x \ \ \iff \ \ 24 x^2-14x+1=0\tag{9}$$
whose roots are
$$x_1=(\cos a_1)^2=1/2 \ \ \text{and} \ \ x_2=(\cos a_2)^2=1/12\tag{10}$$
Therefore, as we are looking for positive values of $\cos a$:
$$\cos a_1=\dfrac{\sqrt{2}}{2} \ \ \text{and} \ \ \cos a_2=\dfrac{1}{2\sqrt{3}}\tag{11}$$
Consequently (see (8)), the two solutions are:
$$BC=10 \cos a_1=5 \sqrt{2}  \ \ \text{and} \ \ BC=10 \cos a_2=\dfrac{5}{\sqrt{3}}\tag{12}$$
Remarks :
a) Figure 1 corresponds to the second case in (12).
b) This solution doesn't use any specific triangle formula.



*Second solution:

Let $a:=AD$; due to chord theorem in circles, we have :
$$DB=6/a \ \ \ \ \& \ \ \ \ EC=4/a\tag{1}$$
Consider Fig. 2.

Fig. 2.
In rectangular triangles $AHD$ and $ADI$, we have :
$$\sin(\tfrac12 A) = \dfrac 1a \ \ \ \ \& \ \ \ \ \tan(\tfrac12 A) = \dfrac{r}{a}\tag{2}$$
Using relationship $1+\dfrac{1}{\tan^2 x}=\dfrac{1}{\sin^2x}$, we deduce from (2) that:
$$r^2=\dfrac{a^2}{a^2-1}\tag{3}$$
Besides, we have this triangle formula:
$$\tan(\tfrac12 A)\tan(\tfrac12 B)\tan(\tfrac12 C)=\dfrac{r}{p}\tag{4}$$
where $p$ is the semiperimeter (half of perimeter).
Relationship (4) can be written:
$$\dfrac{r}{a}\dfrac{r}{(6/a)}\dfrac{r}{(4/a)}=\dfrac{r}{a+\tfrac6a+\tfrac4a}\tag{5}$$
that can be transformed into
$$r^2=\dfrac{24}{a^2+10}\tag{6}$$
Equating (3) and (6) gives a second degree polynomial equation in $x=a^2$:
$$x(x+10)=24(x-1)$$
whose two real positive roots are $x_1=2=a_1^2$ and $x_2=12=a_2^2$.
And we find back the two solutions for $BC=\dfrac{10}{a}$:
$$BC=\dfrac{10}{a_1}=\dfrac{10}{\sqrt{2}} \ \ \ \text{and} \ \ \ BC=\dfrac{10}{a_2}=\dfrac{5}{\sqrt{3}} $$
Remark: relationship (4) could have been obtained using Heron's formula.
