Two circles, with centres O and P respectively, intersect at A and B. The extension of OB intersects the second sircle at C and the extension of PB intersects the first circle at D. A line through B parallel to CD intersects the first circle at Q not equal to B. Prove that AD = BQ.
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We will show that $QD=AB$. Let the extension of $CO$ and $DP$ intersect with their respective circle at $M$ and $N$. It's clear that $M,N,A$ are collinear. So we need to prove that $$\angle QBD=\angle BMA$$ Since $CD$ is parallel to $QB$, it suffices to show that $\square CDMN$ is cyclic which is equivalent to showing that $\square OPCD$ is cyclic. By power of points, $$R(R+BD)=PO^2-r^2$$ , and $$r(r+BC)=PO^2-R^2$$ , where $R,r$ are denoted by the radii of the first and second circle. Therefore $R(BD)=r(BC)$, that is $\square OPDC$ is cyclic. The conclusion follows.
Ahmbak
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thank you for your solution. But could you tell me how did you get the idea, and also please give me some tips on how to deal with such geometry problems. – Nov 24 '15 at 12:30