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How are CD and CT related?

How are QD and QB related?

How are QD and QB related?

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You mean [imath]\triangle[/imath]DQC and [imath]\triangle[/imath]BQA...You seem to be giving us the problem in pieces.

But consider [math]\triangle DCQ[/math] and [math]\triangle AQB[/math].

-Dan

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Please answer these questions specifically. We need to see you actually thinking, in order to direct your thoughts. If some triangles are similar, which ones, and why? (It helps if you use the reply button so we can tell what question you are answering.)How are CD and CT related?

How are QD and QB related?

My first question relates to an important fact about tangents. The second has to do with similar triangles.

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Try thinking about triangle BQT (or CQT). Do you see thatYes, these triangles are similar by angle-angle-angle, DC=CT and TB=ABbut I can't find any connection with TQ.

Either of these approaches is easier to see if you draw some additional line segments, such as the TQ you asked about. Doing so reveals additional triangles to consider.

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AB is perpendicular to AD because ABCD is right trapezoid, but I don't how to prove that TQ is parallel to AB or CD. I will just skip that problem, I've wasted too much time on that proof. Thank you guys

What is the ratio DQ : QB?

What is the ratio CT : TB? (In fact, what

Can you use those to show that QTB is similar to DCB?

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

D(0,1) A(0,-1)

C is on

[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

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.

Q.E.D

If you

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.