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Knowing that Triangle $LAB$ is similar to Triangle $LRQ$, prove that the length of $QR$ is constant while point $L$ varies. There are two circles intersect at points $A$ and $B$. $L$ is a point on first circle that is free to move, whereas $LA$ & $LB$ meet at the second circle again at $Q$ & $R$. $LA$ is not tangent to the second circle.

Should I use proportions from secant segment theorem here to show that $QR$ is not affected by the movement of point $L$? Would that be enough to prove this $QR$ to be constant?

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    @Matthew: I have it written above. Let me know what you think.2012-05-18

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A start: Draw a picture, including line segments $AB$ and $QR$. Note that $\angle LAB + \angle QAB=180^\circ$. But $\angle QAB+\angle BRQ=180^\circ$, since opposite angles of a cyclic quadrilateral add up to $180^\circ$.

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    I was thinking of using the side splitting theorem to show that the ratio of sides were in proportion. LA/AB=PR/PQ. Does that sound like I'm on the right track?2012-05-13
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    Or could I use the angle similarity theorem by proving the pairs of angles congruent by those angles that have intercepted the same arcs within the circles.2012-05-13
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    You could use various theorems, it all depends on what has been proved so far. I used the theorem that says that opposite angles in a quadrilateral inscribed in a circle are supplementary because I was pretty sure that one had been proved already.2012-05-13
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    I never learned about cyclic quadrilaterals. Thanks for your help!2012-05-14
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    @user31284: To prove, there are $2$ cases, centre of the circle inside quadrilateral, or not inside. Proofs quite similar. Let's do centre inside. Draw lines from centre to vertices. We get $4$ isosceles triangles. Call their angles $x$, $x$, $y$, $y$, $z$, $z$, $w$, $w$. Then each sum of opposite angles is $x+y+z+w$, half of $2x+2y+2z+2w$, which is $360^\circ$.2012-05-14
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    Does using inscribed arcs help us? Knowing that the measure of angle LBA is congruent to half of measure arc AL; measure of angle ALB is equal half the measure of arc AB; measure of QRB is equal to half of measure arc BQ.2012-05-17
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    @user31284: Arc calculations can also be used to prove the result. To me, the inscribed quadrilateral stuff is easiest, but there are many approaches.2012-05-17
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    Andre N: I really like your method to use substitution after finding the linear pairs. It is not that easy for me to see the second pair ∠QAB+∠BRQ to sum up to 180 degrees or sum of inscribed angled at up to half the degrees in a circle. Just amazing. I am working on seeing which to prove QR is going to be the same if L moves outside the circle.2012-05-17