Geometry with circles. 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.
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.
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.