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Say we have two circles that intersect at points $X$ and $Y$. We also have a point $A$ on one of those circles, call it the "first circle", with a tangent line $T$ at the point $A$. Lines are extended from $A$ to $X$ and $Y$ until they intersect with the other circle, the "second circle" at points $B$ and $C$ respectively.

My end game is to prove that $BC$ and $T$ are parallel; here's what I have so far:

The tangent $T$ is perpendicular to the radius of the first circle. Extend this radius into a line, call it $K$. The chord $BC$ has a perpendicular bisector that passes through the centre of its circle. Extend this perpendicular bisector into a line, call it $L$.

It's clear that K and L are parallel, but I'm unsure of how to prove this.

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Note that $BCYX$ is a cyclic quadrilateral and hence $$\underbrace{\angle BCY = \angle YXA}_{\text{Exterior angle of a cyclic quadrilateral equals the interior opposite angle}}\\ \underbrace{\angle YXA = \angle TAY}_{\text{External angle of a triangle is equal to the opposite interior angle of the triangle}}$$ Hence, $$\angle BCY = \angle TAY.$$ This means the alternate angle are equal and hence the lines are parallel.

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  • $\begingroup$ That's a much easier way of going about this question. Thanks. $\endgroup$ – Charlie Jun 5 '12 at 4:24
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$$\angle TAY=\angle YXA=\angle BCY$$ Of course, the last equality gets you done as you've proved that alternate interior angles wrt lines $\,TA\,,\,BC\,$ are equal.

Now, how's the last equality proved? Look at the complementary angle of $\,\angle YXA\,\,,\,i.e.\,\,\angle BXY$, which is the opposite angle of a circumscribed quadrilateral in the leftmost circle...

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