Length of Chord is Independent of Point P This is question 1.49 from Baragar's textbook called A Survey of Classical and Modern Geometries if anybody is familiar with that text. This question is assigned as homework, so I am just looking for some places to start, not the full answer. The question goes:
Let two circles $T$ and $T'$ intersect at points $A$ and $B$. Let $P$ be a point on the circle $T$. Let $PA$ intersect $T'$ again at $C$ and let $PB$ intersect $T'$ again at $D$. Show that $|CD|$ is independent of the location of $P$. (Note that there are two cases to consider.)
Now, my professor gave us a hint. He said to connect $A$ to $B$ and look for similar triangles. I am not seeing as to how that helps, the only similar triangles I see would be in the case that $CD$ is parallel to $AB$ and even then we don't know anything about the measures of the triangles so we can't find the exact proportions. Also, I don't know what the two cases we need to consider are. Any help is appreciated.
EDIT: Here is my attempt at the problem so far. I think I have solved one case. First construct segment $AB$. Then there are two cases. The first case is when $ABDC$ forms a quadrilateral. Then this quadrilateral is cyclic (all points lie on $T'$) and thus they have opposite angles that are supplementary (a previous result from the class). This implies angle $CDB$ is congruent to angle $PAB$ and angle $PCD$ is congruent to $ABP$. Thus triangles $ABP$ and $DCP$ are similar. Now we have
$$\frac{|CD|}{|CP|} = \frac{|BA|}{|BP|}$$
$$|CD| = \frac{|BA||CP|}{|BP|}$$
With |BA| being constant and $\frac{|CP|}{|BP|}$ also constant (it is the ratio between the similar side lengths). The second case is pretty similar. I am concerned about the part where I claim $\frac{|CP|}{|BP|}$ is constant, no matter where $P$ is on $T$. Does this need more justification or is it obvious?

Lemma: If the chord subtends an angle θ at the circumference of a circle of given radius (r), then its length is equal to 2r sin θ.
Proof: All basic geometrical knowledge are assumed including (Angle at center = 2 x angle at circumference).
Referring to the figure shown below:-

MN = 2 MK = 2. r sin θ .
After drawing the standard diagram according to the given, we shift P to P’ and draw the rest (as in red) according also to the given. (See the figure above.)
Note that, for some reasons, all angles marked in green are equal, especially α = β
CD = 2 r sin (Ø + β); where r is the radius of the circle CDBA
C’D’ = 2 r sin (Ø + α)
Therefore CD = C’D’
That is, CD is independent of the position of P as long as it lies on the same segment.
