Textbook geometry problem 
Radius of $A$ is $2$, Radius of $B$ is $1$, Radius of $C$ is $4$, Radius of $X$ is $3$. Find the radius of $D$
Could someone please help with this question? :)
EDIT: How I tried to solve it - I tried using similar triangles and other things but to be honest I really don't know how to do this question. This led me to getting 5 as the (incorrect) answer so I'm still not sure what to do really..
However, I found that the tangents of the radii are parallel, if that helps anyone!
Update again: I thought about using the cosine law and setting up two triangles: CX and C's perpendicular radius and CA and C's perpendicular radius
Yet another update: I thought about maybe trying this


Sorry for this bad translation but it's originally in japanese, hopefully you can understand!
The contact of the circle X and QR, E, the contact of the circle D and QR, I and F. (Following contacts, please look at the figure) 
and the inscribed circle X, I explained in Sobase~tsuen D. And inscribed circle, from the nature of the near contact circle, 
(inscribed circle X of radius) × (Sobase~tsuen D of radius) = QE × QF, ※ supplemental reference 
(QE = RF = a, QF = RE = d, You radius x beside contact circle.) 
Thus, ad = 3x = ... ① 
Similarly, inscribed circle X, to think in Sobase~tsuen C, and you have (PH = PG = b,) 
ab = 3 × 4 ... ② 
inscribed circle B, and think in Sobase~tsuen X, (and with SG = SK = c,) 
BC = 1 × 3 ... ③ 
inscribed circle A, to think in Sobase~tsuen X, 
cd = 2 × 3 ... ④ 
than ① × ③ = ② × ④, x = 8 
Therefore, the radius of the circle D is, 8 
(A radius) × (C radius of) = (radius of B) × (radius of D) 
to be, I understand. Supplemental ※ proof, △ XQE∽ △ QDF than (∠XQD = 90 °, more, ∠XQE = ∠QDF) QE: DF = XE: QF than, DF × XE = QE × QF
