In triangle $ABC$ you have the following angles $\angle A=120^\circ$ (top corner), $\angle B=20^\circ$ (left corner) and $\angle C=40^\circ$ (right corner). Denote the central point with $D$ and introduce lengths $AD=a, BD=b,CD=c$.
By law of sines applied to triangles $ABD,ACD,BCD$:
$$a\sin20^\circ=b\sin10^\circ\tag{1}$$
$$a\sin100^\circ=c\sin(40^\circ-x)\tag{2}$$
$$b\sin10^\circ=c\sin x\tag{3}$$
From (1) and (2):
$$b=\frac{a\sin20^\circ}{\sin10^\circ}$$
$$c=\frac{a\sin100^\circ}{\sin(40^\circ-x)}$$
Replace that into (3):
$$\frac{a\sin20^\circ}{\sin10^\circ}\sin10^\circ=\frac{a\sin100^\circ}{\sin(40^\circ-x)}\sin x$$
$$\sin20^\circ \sin(40^\circ-x)=\sin100^\circ\sin x$$
$$\sin20^\circ \sin(40^\circ-x)=\cos10^\circ\sin x$$
$$2\sin10^\circ \cos10^\circ \sin(40^\circ-x)=\cos10^\circ\sin x$$
$$2\sin10^\circ \sin(40^\circ-x)=\sin x$$
Sometimes you have to make things more complicated before your are able to jump over the last hurdle: multiply the right side with $1=2\sin30^\circ$.
$$2\sin10^\circ \sin(40^\circ-x)=2\sin x\sin30^\circ$$
$$\cos(-30^\circ+x)-\cos(50^\circ-x)=\cos(x-30^\circ)-\cos(x+30^\circ)$$
$$\cos(50^\circ-x)=\cos(x+30^\circ)$$
For obviously acute angle $x$
$$50^\circ-x=x+30^\circ$$
$$x=10^\circ$$
No calculator needed.
The formula to be derived are so messy that I can only write down the general ideas and illustrate the work with a special example.
Refer to the figure, let $p=s+t$.
From $\Delta ABX$,
$$ XB^2=AX^2+AB^2-2(AX)(AB)\cos \alpha$$
$$(r+s)^2=r^2+c^2-2rc\cdot \cos \alpha$$
$$r^2+2rs+s^2=r^2+c^2-2rc\cdot \cos \alpha$$
$$2r(s+c \cdot \cos \alpha)=c^2-s^2 \tag{1}$$
Similarly from $\Delta ACX$,
$$2r(p+ b \cdot \cos(A-\alpha))=b^2-p^2 \tag{2}$$
From $(1), (2)$, we have
$$(c^2-s^2)(p+b \cdot \cos(A-\alpha))=(b^2-p^2)(s+c\cdot \cos \alpha)$$
$$(c^2-s^2)(p+b \cdot (\cos A \cos \alpha+ \sin A \sin \alpha))=(b^2-p^2)(s+c\cdot \cos \alpha)$$
The equation above can be reduced to the form:
$$m \cdot \cos \alpha + n \cdot \sin \alpha = k \tag{3} $$
From $(3)$, we can find $\alpha$ by standard method Solving $a \cos x + b \sin x = c$ and hence $r$ from either $(1)$ or $(2)$.
Example: Suppose we want to solve for $r$ with the information given in the figure.
From $\Delta ABX$,
$$(r+\sqrt 3-1)^2=r^2+2^2-4r\cos \alpha$$
$$2r((\sqrt 3 -1)+2 \cdot \cos \alpha) = 4-(\sqrt 3 - 1)^2 $$
$$r((\sqrt 3 -1)+2\cos \alpha) = \sqrt 3 \tag{1}$$
Similarly from $\Delta ACX$,
$$(r+\sqrt {13}-1)^2=r^2+16-8r\cos(120^{\text o}-\alpha)$$
$$2r((\sqrt {13}-1)+4 \cdot \cos (120^{\text o}-\alpha))= 16-(\sqrt {13}-1)^2$$
$$r((\sqrt {13}-1)+4\cos (120^{\text o}-\alpha))= 1+ \sqrt {13} \tag{2}$$
$(1), (2) \implies$
$$\sqrt 3\left( \sqrt {13}-1 +4 \cos(120^{\text o}-\alpha) \right)=(\sqrt {13}+1)\left(\sqrt 3 -1 + 2 \cos \alpha \right)$$
$$\sqrt {13}-2\sqrt 3+1=(2\sqrt{13}+2)\cos \alpha -4\sqrt 3(\cos 120^{\text o} \cos \alpha + \sin 120^{\text o}\sin \alpha )$$
$$(2\sqrt {13}+2\sqrt 3+2)\cos \alpha - 6 \sin \alpha = \sqrt {13}-2\sqrt 3 +1 \tag{3}$$
$(3)$ is of the form
$$m \cdot \cos \alpha + n \cdot \sin \alpha = k $$
which can be solved by standard method.
Indeed the solution of $(3)$ is
$$\alpha = 60^{\text o}$$
Once we have $\alpha = 60^{\text o}$, we can find the value of $r$ from $(1)$
$$r((\sqrt 3 -1)+2\cos \alpha) = \sqrt 3$$
which gives $$r=1.$$
Best Answer
First and last equality are because of triangle similarty ($BQF\sim BDC$ and $AEQ \sim ADC$) and in the middle because of Thales theorem (in angle through $Q$).
$$ {QF \over CD} = {QB\over DB} = {QA\over AC} = {EQ\over CD}$$