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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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    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