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 [imath]x^2+y^2=1[/imath])
Then OT has gradient [imath]\tfrac{y}{x}[/imath] and therefore CB has gradient [imath]-\tfrac{x}{y}[/imath] (assuming, that CB is a tangent at T and therefore perpendicular to OT).
[imath]\therefore[/imath] CB goes through [imath](x,y)[/imath] with gradient [imath]-\tfrac{x}{y}[/imath]
Eqn of
CB: [imath]\boxed{\;\boldsymbol{Y}-y=-\tfrac{x}{y}(\boldsymbol{X}-x)\;}[/imath] (*)
D(0,1) A(0,-1)
C is on
CB (*) and has
Y co-ordinate = 1 [imath]\hspace4ex \therefore \boldsymbol{X}=\tfrac{y(1-y)}{-x}+x[/imath]
[imath]\hspace41ex \boldsymbol{X}=\tfrac{1-y}{x} \hspace6ex[/imath] (since [imath]x^2+y^2=1[/imath])
C[imath]\left(\tfrac{1-y}{x},1\right)[/imath]
Similarly B is on
CB (*) and has
Y co-ordinate = -1[imath]\hspace4ex \rightarrow \boldsymbol{X}=\tfrac{1+y}{x}[/imath]
B [imath]\left(\tfrac{1-y}{x},-1\right)[/imath]
Now equation of AC: A(0,-1) C[imath]\left(\tfrac{1-y}{x},1\right)[/imath]
[imath]\boxed{\,AC\,} \hspace2ex \boldsymbol{Y}=\tfrac{2x}{1-y}(\boldsymbol{X})-1[/imath]
Eqn of DB: D(0,1) B[imath]\left(\tfrac{1+y}{x},-1\right)[/imath]
[imath]\boxed{\,DB\,} \hspace2ex \boldsymbol{Y}=-\tfrac{2x}{1+y}(\boldsymbol{X})+1[/imath]
Q is at the intersection of [imath]\boxed{\,AC\,}, \boxed{\,DB\,}[/imath]
[imath]\rightarrow \left(\tfrac{2x}{1-y}+\tfrac{2x}{1+y}\right)\boldsymbol{X}=2[/imath]
[imath]\tfrac{4x}{1-y^2} \boldsymbol{X}=2[/imath]
[imath]\boldsymbol{X}=\tfrac{x}{2} \hspace10ex[/imath] note [imath]1-y^2=x^2[/imath]
[imath]\therefore \boldsymbol{Y}=y \hspace8ex[/imath] using e.g.
DB, note [imath]1-x^2=y^2[/imath]
Q.E.D
If you
do want to do this using similar triangles:
Assuming CB is tangent to the circle at T, then:
[imath]\bigtriangleup OTU \backsim \bigtriangleup TSU \backsim \bigtriangleup TBV[/imath]
[imath]\hspace10ex \bigtriangleup TSU \backsim \bigtriangleup TCR \hspace5ex (\backsim\;[/imath] similar[imath])[/imath]
[imath]TU=y, US=\tfrac{y}{x} . y = \tfrac{y^2}{x}, VB=\tfrac{1+y}{y} . \tfrac{y^2}{x}=\tfrac{y+y^2}{x}\\
\hspace15ex US=\tfrac{y^2}{x}, RC=\tfrac{1-y}{y} . \tfrac{y^2}{x} = \tfrac{y-y^2}{x}[/imath]
[imath]DC = DR - RC[/imath]
[imath]=x-\tfrac{y-y^2}{x}\\
\boxed{\;DC=\tfrac{1-y}{x}\;} \hspace5ex[/imath] (since [imath]x^2+y^2=1[/imath])
[imath]AB = AV + VB = x+ \tfrac{y+y^2}{x}\\
\boxed{\;AB= \tfrac{1+y}{x}\;}[/imath]
Assuming CD and AB are tangents at D and A respectively, then CD and AB are parallel, therefore [imath]\bigtriangleup DCQ \backsim \bigtriangleup BAQ[/imath]
Now [imath]AB =\tfrac{1+y}{1-y} DC\\
\therefore QB=\tfrac{1+y}{1-y} DQ\\
\therefore DB = DQ + \tfrac{1+y}{1-y} DQ\\
DB = \tfrac{2}{1-y} DQ[/imath]
Now [imath]\bigtriangleup DWQ \backsim \bigtriangleup DAB, \hspace3ex \text{ and from previous line s.f. } \tfrac{1-y}{2}[/imath] (to get sides in [imath]\bigtriangleup DWQ[/imath])
[imath]\therefore WQ = \tfrac{1-y}{2} . AB = \tfrac{1-y}{2} . \tfrac{1+y}{x}=\tfrac{1-y^2}{2x}=\tfrac{x^2}{2x}=\tfrac{x}{2}[/imath]
[imath]DW = \tfrac{1-y}{2} DA = \tfrac{1-y}{2} . 2 = 1-y[/imath]
i.e. [imath]WO=y[/imath]
Now [imath]\hspace1ex QO=WO[/imath]
[imath]\therefore QO=y[/imath]
[imath]\therefore[/imath] Q is at the same height as T.
Q.E.D.