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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?

ruferd
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2 Answers2

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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:- enter image description here

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.

Mick
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  • Thank you for you reply! It seems very easy once we have that Lemma you pointed out, which is something we haven't seen in my course yet. I would vote up both of the responses but I don't have reputation just yet. – ruferd Sep 28 '13 at 20:28
  • @ruferd Voting up is minor, and the important point is the solution helpful to you. – Mick Sep 29 '13 at 06:35
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Hint: The angle $APB$ is fixed for all $P$ (facing arc $AB$ in $T$) so the angles $PCD$ and $ P'C'D'$ are the same for all $P$ and $P'$. This angle is also determined by difference of two arcs $AB$ and $CD$ in $T'$. You can now solve the problem.

Arash
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  • Thank you for a quick reply. Yes, this solution does make sense to me, and it seems quite short and sweet. My concern however is that so far in our course, we have not discussed measuring arcs very much. I think he is aiming for an approach that utilizes similar triangles. I could use your method, but then I would have to introduce the idea of measuring arcs that are formed by angles that are outside of the circle. – ruferd Sep 21 '13 at 21:43