Two circles intersect in points $A$ and $B$.$PQ$ is a line segment through $A$ and terminating on the two circles. Prove that $\frac{BP}{BQ}$ is constant for all allowable configuration of PQ.
May be it is a easy question, but I failed to answer it.I tried to rotate the small circle along the line $AB$.Please give me some hints.
Let $R_1$ and $R_2$ be radii of our circles.
Then your statement follows immediately from the law of sines.
The hint for another way without trigonometry.
Let $O_1$ and $O_2$ be centers of circles $(PAP)$ and $QAB$)$ respectively
and let $M$ and $N$ be midpoints of $PB$ and $QB$ respectively.
Prove that $\Delta PMO_1\sim\Delta QNO_2$.
This was a similar triangles problem.
Firstly, let $O_1$ be the centre of the circle on which $P$ is on and $O_2$ be the centre of the circle on which $Q$ is on.
It is easy to show $O_1 O_2$ bisects $AB$. Thus, $\angle BO_1 O_2 = \frac{1}{2} BO_1 A$.
Also, $\angle BO_1 A = 2 \angle BPA$. Thus, $\angle BO_1 O_2 = \angle BPA$.
Similarly, we can show $\angle BO_2 O_1 = \angle BQA$.
Thus, $\Delta BO_1 O_2 \sim BPQ$ since they are equiangular. Therefore, $\frac{BP}{BQ}=\frac{BO_1}{BO_2}$, which always remains constant.
Thus, proven.
