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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. – user31284 May 18 '12 at 3:51
up vote 1 down vote accepted

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? – user31284 May 13 '12 at 22:24
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. – user31284 May 13 '12 at 22:42
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. – André Nicolas May 13 '12 at 23:21
I never learned about cyclic quadrilaterals. Thanks for your help! – user31284 May 14 '12 at 3:25
@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 $360^\circ$. – André Nicolas May 14 '12 at 4:02

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