It may be more straightforward to deal with the equations of the lines rather than similar triangles.
Taking AD to be vertical, all we have to show is that Q has the same Y co-ordinate as T.
Take the radius of the circle to be 1 (wlog), O (0,0) and T has co-ordinates (x,y)
(Note
x2+y2=1)
Then OT has gradient
xy and therefore CB has gradient
−yx (assuming, that CB is a tangent at T and therefore perpendicular to OT).
∴ CB goes through
(x,y) with gradient
−yx
Eqn of
CB:
Y−y=−yx(X−x) (*)
D(0,1) A(0,-1)
C is on
CB (*) and has
Y co-ordinate = 1
∴X=−xy(1−y)+x
X=x1−y (since
x2+y2=1)
C
(x1−y,1)
Similarly B is on
CB (*) and has
Y co-ordinate = -1
→X=x1+y
B
(x1−y,−1)
Now equation of AC: A(0,-1) C
(x1−y,1)
ACY=1−y2x(X)−1
Eqn of DB: D(0,1) B
(x1+y,−1)
DBY=−1+y2x(X)+1
Q is at the intersection of
AC,DB
→(1−y2x+1+y2x)X=2
1−y24xX=2
X=2x note
1−y2=x2
∴Y=y using e.g.
DB, note
1−x2=y2
Q.E.D
If you
do want to do this using similar triangles:

Assuming CB is tangent to the circle at T, then:
△OTU∽△TSU∽△TBV
△TSU∽△TCR(∽ similar
)
TU=y,US=xy.y=xy2,VB=y1+y.xy2=xy+y2US=xy2,RC=y1−y.xy2=xy−y2
DC=DR−RC
=x−xy−y2DC=x1−y (since
x2+y2=1)
AB=AV+VB=x+xy+y2AB=x1+y
Assuming CD and AB are tangents at D and A respectively, then CD and AB are parallel, therefore
△DCQ∽△BAQ
Now
AB=1−y1+yDC∴QB=1−y1+yDQ∴DB=DQ+1−y1+yDQDB=1−y2DQ
Now
△DWQ∽△DAB, and from previous line s.f. 21−y (to get sides in
△DWQ)
∴WQ=21−y.AB=21−y.x1+y=2x1−y2=2xx2=2x
DW=21−yDA=21−y.2=1−y
i.e.
WO=y
Now
QO=WO
∴QO=y
∴ Q is at the same height as T.
Q.E.D.